An antiderivative of f f f F F F F ′ = f F' = f F ′ = f indefinite integral is:
Definite Integral as Area Under a Curve
Adjust the parameters in the graph above to explore the relationships between variables.
∫ f ( x ) d x = F ( x ) + C \int f(x)\, dx = F(x) + C ∫ f ( x ) d x = F ( x ) + C Where C C C
| Function f ( x ) f(x) f ( x ) F ( x ) F(x) F ( x ) x n x^n x n n ≠ − 1 n \ne -1 n = − 1 x n + 1 n + 1 + C \frac{x^{n+1}}{n+1} + C n + 1 x n + 1 + C 1 x \frac{1}{x} x 1 ln ∣ x ∣ + C \ln | x | + C ln ∣ x ∣ + C e x e^x e x e x + C e^x + C e x + C sin x \sin x sin x − cos x + C -\cos x + C − cos x + C cos x \cos x cos x sin x + C \sin x + C sin x + C sec 2 x \sec^2 x sec 2 x tan x + C \tan x + C tan x + C csc 2 x \csc^2 x csc 2 x − cot x + C -\cot x + C − cot x + C sec x tan x \sec x \tan x sec x tan x sec x + C \sec x + C sec x + C csc x cot x \csc x \cot x csc x cot x − csc x + C -\csc x + C − csc x + C 1 1 + x 2 \frac{1}{1+x^2} 1 + x 2 1 arctan x + C \arctan x + C arctan x + C 1 1 − x 2 \frac{1}{\sqrt{1-x^2}} 1 − x 2 1 arcsin x + C \arcsin x + C arcsin x + C
∫ x n d x = x n + 1 n + 1 + C , n ≠ − 1 \int x^n\, dx = \frac{x^{n+1}}{n+1} + C, \quad n \ne -1 ∫ x n d x = n + 1 x n + 1 + C , n = − 1 Why the power rule excludes n = − 1 n = -1 n = − 1 Substituting n = − 1 n = -1 n = − 1 x 0 0 \frac{x^0}{0} 0 x 0 1 x \frac{1}{x} x 1 ln ∣ x ∣ \ln|x| ln ∣ x ∣
Why the absolute value in ln ∣ x ∣ \ln|x| ln ∣ x ∣ The derivative of ln x \ln x ln x 1 x \frac{1}{x} x 1 x > 0 x \gt 0 x > 0 x < 0 x \lt 0 x < 0 ln ( − x ) \ln(-x) ln ( − x ) 1 − x ⋅ ( − 1 ) = 1 x \frac{1}{-x} \cdot (-1) = \frac{1}{x} − x 1 ⋅ ( − 1 ) = x 1 1 x \frac{1}{x} x 1 0 0 0 ln ∣ x ∣ + C \ln|x| + C ln ∣ x ∣ + C
Example
Evaluate ∫ ( 3 x 4 − 2 x 2 + 5 x − 1 ) d x \displaystyle\int (3x^4 - 2x^2 + 5x - 1)\, dx ∫ ( 3 x 4 − 2 x 2 + 5 x − 1 ) d x
∫ ( 3 x 4 − 2 x 2 + 5 x − 1 ) d x = 3 x 5 5 − 2 x 3 3 + 5 x 2 2 − x + C \int (3x^4 - 2x^2 + 5x - 1)\, dx = \frac{3x^5}{5} - \frac{2x^3}{3} + \frac{5x^2}{2} - x + C ∫ ( 3 x 4 − 2 x 2 + 5 x − 1 ) d x = 5 3 x 5 − 3 2 x 3 + 2 5 x 2 − x + C Example
Evaluate ∫ 3 x 2 d x \displaystyle\int \frac{3}{x^2}\, dx ∫ x 2 3 d x
Rewrite as ∫ 3 x − 2 d x = 3 x − 1 − 1 + C = − 3 x + C \displaystyle\int 3x^{-2}\, dx = \frac{3x^{-1}}{-1} + C = -\frac{3}{x} + C ∫ 3 x − 2 d x = − 1 3 x − 1 + C = − x 3 + C
Example
Evaluate ∫ 2 x 3 − x + 4 x d x \displaystyle\int \frac{2x^3 - x + 4}{\sqrt{x}}\, dx ∫ x 2 x 3 − x + 4 d x
Rewrite: ∫ ( 2 x 5 / 2 − x 1 / 2 + 4 x − 1 / 2 ) d x = 4 x 7 / 2 7 − 2 x 3 / 2 3 + 8 x 1 / 2 + C \displaystyle\int (2x^{5/2} - x^{1/2} + 4x^{-1/2})\, dx = \frac{4x^{7/2}}{7} - \frac{2x^{3/2}}{3} + 8x^{1/2} + C ∫ ( 2 x 5/2 − x 1/2 + 4 x − 1/2 ) d x = 7 4 x 7/2 − 3 2 x 3/2 + 8 x 1/2 + C
A Riemann sum approximates the area under a curve by dividing the region into rectangles:
∑ i = 1 n f ( x i ∗ ) Δ x \sum_{i=1}^{n} f(x_i^*) \Delta x i = 1 ∑ n f ( x i ∗ ) Δ x Where Δ x = b − a n \Delta x = \frac{b - a}{n} Δ x = n b − a x i ∗ x_i^* x i ∗ i i i
Type Sample Point x i ∗ x_i^* x i ∗ Left Riemann sum Left endpoint Right Riemann sum Right endpoint Midpoint sum Midpoint of subinterval Trapezoidal sum Average of endpoints (trapezoids, not rectangles)
Why Riemann sums matter. They are the foundation of the definite integral. As n → ∞ n \to \infty n → ∞
Theorem. If f f f [ a , b ] [a, b] [ a , b ] x i ∗ x_i^* x i ∗
The definite integral of f f f a a a b b b
∫ a b f ( x ) d x = lim n → ∞ ∑ i = 1 n f ( x i ∗ ) Δ x \int_a^b f(x)\, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x ∫ a b f ( x ) d x = n → ∞ lim i = 1 ∑ n f ( x i ∗ ) Δ x Provided this limit exists. When it does, f f f integrable on [ a , b ] [a, b] [ a , b ]
∫ a a f ( x ) d x = 0 \displaystyle\int_a^a f(x)\, dx = 0 ∫ a a f ( x ) d x = 0 ∫ a b f ( x ) d x = − ∫ b a f ( x ) d x \displaystyle\int_a^b f(x)\, dx = -\int_b^a f(x)\, dx ∫ a b f ( x ) d x = − ∫ b a f ( x ) d x ∫ a b [ f ( x ) ± g ( x ) ] d x = ∫ a b f ( x ) d x ± ∫ a b g ( x ) d x \displaystyle\int_a^b [f(x) \pm g(x)]\, dx = \int_a^b f(x)\, dx \pm \int_a^b g(x)\, dx ∫ a b [ f ( x ) ± g ( x )] d x = ∫ a b f ( x ) d x ± ∫ a b g ( x ) d x ∫ a b c ⋅ f ( x ) d x = c ∫ a b f ( x ) d x \displaystyle\int_a^b c \cdot f(x)\, dx = c \int_a^b f(x)\, dx ∫ a b c ⋅ f ( x ) d x = c ∫ a b f ( x ) d x ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x \displaystyle\int_a^b f(x)\, dx = \int_a^c f(x)\, dx + \int_c^b f(x)\, dx ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x Property 5 (Additivity) is powerful. It allows splitting integrals at discontinuities. For Example, if f f f c ∈ ( a , b ) c \in (a, b) c ∈ ( a , b )
∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x \int_a^b f(x)\, dx = \int_a^c f(x)\, dx + \int_c^b f(x)\, dx ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x Comparison properties. If f ( x ) ≥ g ( x ) f(x) \ge g(x) f ( x ) ≥ g ( x ) [ a , b ] [a, b] [ a , b ] ∫ a b f ( x ) d x ≥ ∫ a b g ( x ) d x \displaystyle\int_a^b f(x)\, dx \ge \int_a^b g(x)\, dx ∫ a b f ( x ) d x ≥ ∫ a b g ( x ) d x
In particular, if m ≤ f ( x ) ≤ M m \le f(x) \le M m ≤ f ( x ) ≤ M [ a , b ] [a, b] [ a , b ] m ( b − a ) ≤ ∫ a b f ( x ) d x ≤ M ( b − a ) m(b-a) \le \displaystyle\int_a^b f(x)\, dx \le M(b-a) m ( b − a ) ≤ ∫ a b f ( x ) d x ≤ M ( b − a )
If f ( x ) ≥ 0 f(x) \ge 0 f ( x ) ≥ 0 [ a , b ] [a, b] [ a , b ] ∫ a b f ( x ) d x \displaystyle\int_a^b f(x)\, dx ∫ a b f ( x ) d x f f f net (signed) area.
The total area between f f f x x x [ a , b ] [a, b] [ a , b ]
\mathrm{Total Area = \int_a^b |f(x)|\, dx Example
Find the total area between f ( x ) = x 2 − 4 f(x) = x^2 - 4 f ( x ) = x 2 − 4 x x x [ − 3 , 3 ] [-3, 3] [ − 3 , 3 ]
Find the zeros: x 2 − 4 = 0 ⟹ x = ± 2 x^2 - 4 = 0 \implies x = \pm 2 x 2 − 4 = 0 ⟹ x = ± 2
\mathrm{Total Area = \int_{-3}^{-2} (x^2 - 4)\, dx + \int_{-2}^{2} (4 - x^2)\, dx + \int_{2}^{3} (x^2 - 4)\, dx = [ x 3 3 − 4 x ] − 3 − 2 + [ 4 x − x 3 3 ] − 2 2 + [ x 3 3 − 4 x ] 2 3 = \left[\frac{x^3}{3} - 4x\right]_{-3}^{-2} + \left[4x - \frac{x^3}{3}\right]_{-2}^{2} + \left[\frac{x^3}{3} - 4x\right]_{2}^{3} = [ 3 x 3 − 4 x ] − 3 − 2 + [ 4 x − 3 x 3 ] − 2 2 + [ 3 x 3 − 4 x ] 2 3 = − 8 − ( − 27 ) 3 − − 8 3 + 8 − ( − 8 ) 3 + 27 − 8 3 = 19 3 + 16 3 + 19 3 = 54 3 = 18 = \frac{-8 - (-27)}{3} - \frac{-8}{3} + \frac{8 - (-8)}{3} + \frac{27 - 8}{3} = \frac{19}{3} + \frac{16}{3} + \frac{19}{3} = \frac{54}{3} = 18 = 3 − 8 − ( − 27 ) − 3 − 8 + 3 8 − ( − 8 ) + 3 27 − 8 = 3 19 + 3 16 + 3 19 = 3 54 = 18 If f f f [ a , b ] [a, b] [ a , b ] g g g
G ( x ) = ∫ a x f ( t ) d t G(x) = \int_a^x f(t)\, dt G ( x ) = ∫ a x f ( t ) d t Is differentiable on ( a , b ) (a, b) ( a , b )
G ′ ( x ) = f ( x ) G'(x) = f(x) G ′ ( x ) = f ( x ) More generally, by the chain rule:
d d x [ ∫ a u ( x ) f ( t ) d t ] = f ( u ( x ) ) ⋅ u ′ ( x ) \frac{d}{dx}\!\left[\int_a^{u(x)} f(t)\, dt\right] = f(u(x)) \cdot u'(x) d x d [ ∫ a u ( x ) f ( t ) d t ] = f ( u ( x )) ⋅ u ′ ( x ) Intuition. FTC Part 1 says: the rate at which the accumulated area changes is just the height of The curve at that point. This connects the two halves of calculus: the derivative and the integral Are inverse operations.
Example
Find d d x [ ∫ 1 x 2 sin ( t 2 ) d t ] \displaystyle\frac{d}{dx}\!\left[\int_1^{x^2} \sin(t^2)\, dt\right] d x d [ ∫ 1 x 2 sin ( t 2 ) d t ]
By FTC Part 1 and the chain rule:
d d x [ ∫ 1 x 2 sin ( t 2 ) d t ] = sin ( ( x 2 ) 2 ) ⋅ 2 x = 2 x sin ( x 4 ) \frac{d}{dx}\!\left[\int_1^{x^2} \sin(t^2)\, dt\right] = \sin\!\left((x^2)^2\right) \cdot 2x = 2x \sin(x^4) d x d [ ∫ 1 x 2 sin ( t 2 ) d t ] = sin ( ( x 2 ) 2 ) ⋅ 2 x = 2 x sin ( x 4 ) Example
Find d d x [ ∫ 0 x e t 2 d t ] \displaystyle\frac{d}{dx}\!\left[\int_0^{\sqrt{x}} e^{t^2}\, dt\right] d x d [ ∫ 0 x e t 2 d t ]
d d x [ ∫ 0 x e t 2 d t ] = e ( x ) 2 ⋅ 1 2 x = e x 2 x \frac{d}{dx}\!\left[\int_0^{\sqrt{x}} e^{t^2}\, dt\right] = e^{(\sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}} = \frac{e^x}{2\sqrt{x}} d x d [ ∫ 0 x e t 2 d t ] = e ( x ) 2 ⋅ 2 x 1 = 2 x e x Example
Find d d x [ ∫ x x 2 1 1 + t 2 d t ] \displaystyle\frac{d}{dx}\!\left[\int_x^{x^2} \frac{1}{1+t^2}\, dt\right] d x d [ ∫ x x 2 1 + t 2 1 d t ]
Split the integral at a constant (say 0 0 0
∫ x x 2 1 1 + t 2 d t = ∫ 0 x 2 1 1 + t 2 d t − ∫ 0 x 1 1 + t 2 d t \int_x^{x^2} \frac{1}{1+t^2}\, dt = \int_0^{x^2} \frac{1}{1+t^2}\, dt - \int_0^x \frac{1}{1+t^2}\, dt ∫ x x 2 1 + t 2 1 d t = ∫ 0 x 2 1 + t 2 1 d t − ∫ 0 x 1 + t 2 1 d t Differentiating:
1 1 + x 4 ⋅ 2 x − 1 1 + x 2 = 2 x 1 + x 4 − 1 1 + x 2 \frac{1}{1+x^4} \cdot 2x - \frac{1}{1+x^2} = \frac{2x}{1+x^4} - \frac{1}{1+x^2} 1 + x 4 1 ⋅ 2 x − 1 + x 2 1 = 1 + x 4 2 x − 1 + x 2 1 If f f f [ a , b ] [a, b] [ a , b ] F F F f f f
∫ a b f ( x ) d x = F ( b ) − F ( a ) \int_a^b f(x)\, dx = F(b) - F(a) ∫ a b f ( x ) d x = F ( b ) − F ( a ) This is the evaluation theorem. We write F ( x ) ∣ a b F(x)\Big|_a^b F ( x ) a b F ( b ) − F ( a ) F(b) - F(a) F ( b ) − F ( a )
Example
Evaluate ∫ 1 3 ( x 2 + 1 x ) d x \displaystyle\int_1^3 \left(x^2 + \frac{1}{x}\right) dx ∫ 1 3 ( x 2 + x 1 ) d x
∫ 1 3 ( x 2 + 1 x ) d x = [ x 3 3 + ln x ] 1 3 = ( 27 3 + ln 3 ) − ( 1 3 + 0 ) = 26 3 + ln 3 \int_1^3 \left(x^2 + \frac{1}{x}\right) dx = \left[\frac{x^3}{3} + \ln x\right]_1^3 = \left(\frac{27}{3} + \ln 3\right) - \left(\frac{1}{3} + 0\right) = \frac{26}{3} + \ln 3 ∫ 1 3 ( x 2 + x 1 ) d x = [ 3 x 3 + ln x ] 1 3 = ( 3 27 + ln 3 ) − ( 3 1 + 0 ) = 3 26 + ln 3 Let F F F f f f g ( x ) = ∫ a x f ( t ) d t g(x) = \int_a^x f(t)\, dt g ( x ) = ∫ a x f ( t ) d t
By FTC Part 1, g ′ ( x ) = f ( x ) = F ′ ( x ) g'(x) = f(x) = F'(x) g ′ ( x ) = f ( x ) = F ′ ( x ) g ( x ) = F ( x ) + C g(x) = F(x) + C g ( x ) = F ( x ) + C C C C
At x = a x = a x = a g ( a ) = 0 = F ( a ) + C ⟹ C = − F ( a ) g(a) = 0 = F(a) + C \implies C = -F(a) g ( a ) = 0 = F ( a ) + C ⟹ C = − F ( a )
Therefore, g ( x ) = F ( x ) − F ( a ) g(x) = F(x) - F(a) g ( x ) = F ( x ) − F ( a )
∫ a b f ( t ) d t = g ( b ) = F ( b ) − F ( a ) \int_a^b f(t)\, dt = g(b) = F(b) - F(a) ∫ a b f ( t ) d t = g ( b ) = F ( b ) − F ( a ) ■ \blacksquare ■
If ∫ f ( g ( x ) ) ⋅ g ′ ( x ) d x \displaystyle\int f(g(x)) \cdot g'(x)\, dx ∫ f ( g ( x )) ⋅ g ′ ( x ) d x u = g ( x ) u = g(x) u = g ( x ) d u = g ′ ( x ) d x du = g'(x)\, dx d u = g ′ ( x ) d x
∫ f ( u ) d u = F ( u ) + C = F ( g ( x ) ) + C \int f(u)\, du = F(u) + C = F(g(x)) + C ∫ f ( u ) d u = F ( u ) + C = F ( g ( x )) + C Strategy for choosing u u u Look for a function and its derivative in the integrand. If you can Spot f ( g ( x ) ) f(g(x)) f ( g ( x )) g ′ ( x ) g'(x) g ′ ( x ) u = g ( x ) u = g(x) u = g ( x )
Example
Evaluate ∫ 2 x e x 2 d x \displaystyle\int 2x e^{x^2}\, dx ∫ 2 x e x 2 d x
Let u = x 2 u = x^2 u = x 2 d u = 2 x d x du = 2x\, dx d u = 2 x d x
∫ 2 x e x 2 d x = ∫ e u d u = e u + C = e x 2 + C \int 2x e^{x^2}\, dx = \int e^u\, du = e^u + C = e^{x^2} + C ∫ 2 x e x 2 d x = ∫ e u d u = e u + C = e x 2 + C Example
Evaluate ∫ x x 2 + 1 d x \displaystyle\int \frac{x}{x^2 + 1}\, dx ∫ x 2 + 1 x d x
Let u = x 2 + 1 u = x^2 + 1 u = x 2 + 1 d u = 2 x d x du = 2x\, dx d u = 2 x d x 1 2 d u = x d x \frac{1}{2}du = x\, dx 2 1 d u = x d x
∫ x x 2 + 1 d x = 1 2 ∫ 1 u d u = 1 2 ln ∣ u ∣ + C = 1 2 ln ( x 2 + 1 ) + C \int \frac{x}{x^2 + 1}\, dx = \frac{1}{2}\int \frac{1}{u}\, du = \frac{1}{2}\ln|u| + C = \frac{1}{2}\ln(x^2 + 1) + C ∫ x 2 + 1 x d x = 2 1 ∫ u 1 d u = 2 1 ln ∣ u ∣ + C = 2 1 ln ( x 2 + 1 ) + C Example
Evaluate ∫ ln x x d x \displaystyle\int \frac{\ln x}{x}\, dx ∫ x ln x d x
Let u = ln x u = \ln x u = ln x d u = 1 x d x du = \frac{1}{x}\, dx d u = x 1 d x
∫ ln x x d x = ∫ u d u = u 2 2 + C = ( ln x ) 2 2 + C \int \frac{\ln x}{x}\, dx = \int u\, du = \frac{u^2}{2} + C = \frac{(\ln x)^2}{2} + C ∫ x ln x d x = ∫ u d u = 2 u 2 + C = 2 ( ln x ) 2 + C When using u u u
∫ a b f ( g ( x ) ) g ′ ( x ) d x = ∫ g ( a ) g ( b ) f ( u ) d u \int_a^b f(g(x))g'(x)\, dx = \int_{g(a)}^{g(b)} f(u)\, du ∫ a b f ( g ( x )) g ′ ( x ) d x = ∫ g ( a ) g ( b ) f ( u ) d u Example
Evaluate ∫ 0 1 x 1 + x 2 d x \displaystyle\int_0^1 x\sqrt{1 + x^2}\, dx ∫ 0 1 x 1 + x 2 d x
Let u = 1 + x^2$$du = 2x\, dx . When x = 0$$u = 1 . When x = 1 , u = 2 x = 1, u = 2 x = 1 , u = 2
∫ 0 1 x 1 + x 2 d x = 1 2 ∫ 1 2 u d u = 1 2 [ 2 u 3 / 2 3 ] 1 2 = 1 3 ( 2 2 − 1 ) \int_0^1 x\sqrt{1 + x^2}\, dx = \frac{1}{2}\int_1^2 \sqrt{u}\, du = \frac{1}{2}\left[\frac{2u^{3/2}}{3}\right]_1^2 = \frac{1}{3}(2\sqrt{2} - 1) ∫ 0 1 x 1 + x 2 d x = 2 1 ∫ 1 2 u d u = 2 1 [ 3 2 u 3/2 ] 1 2 = 3 1 ( 2 2 − 1 ) Example
Evaluate ∫ 0 1 2 x 1 + x 2 d x \displaystyle\int_0^1 \frac{2x}{\sqrt{1 + x^2}}\, dx ∫ 0 1 1 + x 2 2 x d x
Let u = 1 + x^2$$du = 2x\, dx . When x = 0$$u = 1 . When x = 1$$u = 2 .
∫ 0 1 2 x 1 + x 2 d x = ∫ 1 2 u − 1 / 2 d u = [ 2 u ] 1 2 = 2 2 − 2 \int_0^1 \frac{2x}{\sqrt{1 + x^2}}\, dx = \int_1^2 u^{-1/2}\, du = \left[2\sqrt{u}\right]_1^2 = 2\sqrt{2} - 2 ∫ 0 1 1 + x 2 2 x d x = ∫ 1 2 u − 1/2 d u = [ 2 u ] 1 2 = 2 2 − 2 ∫ u d v = u v − ∫ v d u \int u\, dv = uv - \int v\, du ∫ u d v = uv − ∫ v d u Choose u u u LIATE priority: Logarithmic, Inverse trig, Algebraic, Trig, Exponential.
Why LIATE works. The antiderivative of u u u u u u
Tabular integration (DI method). For integrals of the form ∫ f ( x ) g ( x ) d x \displaystyle\int f(x)g(x)\, dx ∫ f ( x ) g ( x ) d x f ( x ) f(x) f ( x ) g ( x ) g(x) g ( x ) f f f g g g +$$-$$+$$- . The result is the sum of diagonal products.
Example
Evaluate ∫ x e x d x \displaystyle\int x e^x\, dx ∫ x e x d x
Let u = x$$dv = e^x\, dx . Then du = dx$$v = e^x .
∫ x e x d x = x e x − ∫ e x d x = x e x − e x + C = e x ( x − 1 ) + C \int x e^x\, dx = xe^x - \int e^x\, dx = xe^x - e^x + C = e^x(x - 1) + C ∫ x e x d x = x e x − ∫ e x d x = x e x − e x + C = e x ( x − 1 ) + C Example
Evaluate ∫ x 2 e x d x \displaystyle\int x^2 e^x\, dx ∫ x 2 e x d x
Let u = x^2$$dv = e^x\, dx . Then du = 2x\, dx$$v = e^x .
= x 2 e x − ∫ 2 x e x d x = x^2 e^x - \int 2x e^x\, dx = x 2 e x − ∫ 2 x e x d x Apply integration by parts again for ∫ 2 x e x d x \int 2x e^x\, dx ∫ 2 x e x d x u = 2x$$dv = e^x\, dx du = 2\, dx$$v = e^x :
∫ 2 x e x d x = 2 x e x − 2 e x + C \int 2x e^x\, dx = 2xe^x - 2e^x + C ∫ 2 x e x d x = 2 x e x − 2 e x + C Therefore:
∫ x 2 e x d x = x 2 e x − 2 x e x + 2 e x + C = e x ( x 2 − 2 x + 2 ) + C \int x^2 e^x\, dx = x^2 e^x - 2xe^x + 2e^x + C = e^x(x^2 - 2x + 2) + C ∫ x 2 e x d x = x 2 e x − 2 x e x + 2 e x + C = e x ( x 2 − 2 x + 2 ) + C Tabular method check:
Sign D (derivatives) I (integrals) + + + x 2 x^2 x 2 e x e^x e x − - − 2 x 2x 2 x e x e^x e x + + + 2 2 2 e x e^x e x
Result: x 2 e x − 2 x e x + 2 e x = e x ( x 2 − 2 x + 2 ) x^2 e^x - 2xe^x + 2e^x = e^x(x^2 - 2x + 2) x 2 e x − 2 x e x + 2 e x = e x ( x 2 − 2 x + 2 )
Example
Evaluate ∫ ln x d x \displaystyle\int \ln x\, dx ∫ ln x d x
Let u = \ln x$$dv = dx . Then du = \frac{1}{x}dx$$v = x .
∫ ln x d x = x ln x − ∫ x ⋅ 1 x d x = x ln x − x + C \int \ln x\, dx = x\ln x - \int x \cdot \frac{1}{x}\, dx = x\ln x - x + C ∫ ln x d x = x ln x − ∫ x ⋅ x 1 d x = x ln x − x + C Example
Evaluate ∫ 0 π / 2 x sin x d x \displaystyle\int_0^{\pi/2} x\sin x\, dx ∫ 0 π /2 x sin x d x
Let u = x$$dv = \sin x\, dx$$du = dx$$v = -\cos x .
= [ − x cos x ] 0 π / 2 − ∫ 0 π / 2 ( − cos x ) d x = ( 0 + π 2 cos 0 ) − [ − sin x ] 0 π / 2 = π 2 − 1 = [-x\cos x]_0^{\pi/2} - \int_0^{\pi/2} (-\cos x)\, dx = \left(0 + \frac{\pi}{2}\cos 0\right) - [-\sin x]_0^{\pi/2} = \frac{\pi}{2} - 1 = [ − x cos x ] 0 π /2 − ∫ 0 π /2 ( − cos x ) d x = ( 0 + 2 π cos 0 ) − [ − sin x ] 0 π /2 = 2 π − 1 Decompose a rational function into simpler fractions before integrating.
Example
Evaluate ∫ 1 x 2 − 1 d x \displaystyle\int \frac{1}{x^2 - 1}\, dx ∫ x 2 − 1 1 d x
Factor: x 2 − 1 = ( x − 1 ) ( x + 1 ) x^2 - 1 = (x-1)(x+1) x 2 − 1 = ( x − 1 ) ( x + 1 )
1 ( x − 1 ) ( x + 1 ) = A x − 1 + B x + 1 \frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} ( x − 1 ) ( x + 1 ) 1 = x − 1 A + x + 1 B 1 = A ( x + 1 ) + B ( x − 1 ) 1 = A(x+1) + B(x-1) 1 = A ( x + 1 ) + B ( x − 1 ) At x = 1 x = 1 x = 1 1 = 2 A ⟹ A = 1 2 1 = 2A \implies A = \frac{1}{2} 1 = 2 A ⟹ A = 2 1 x = − 1 x = -1 x = − 1 1 = − 2 B ⟹ B = − 1 2 1 = -2B \implies B = -\frac{1}{2} 1 = − 2 B ⟹ B = − 2 1
∫ 1 x 2 − 1 d x = 1 2 ∫ 1 x − 1 d x − 1 2 ∫ 1 x + 1 d x = 1 2 ln ∣ x − 1 ∣ − 1 2 ln ∣ x + 1 ∣ + C = 1 2 ln ∣ x − 1 x + 1 ∣ + C \int \frac{1}{x^2 - 1}\, dx = \frac{1}{2}\int \frac{1}{x-1}\, dx - \frac{1}{2}\int \frac{1}{x+1}\, dx = \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C = \frac{1}{2}\ln\!\left|\frac{x-1}{x+1}\right| + C ∫ x 2 − 1 1 d x = 2 1 ∫ x − 1 1 d x − 2 1 ∫ x + 1 1 d x = 2 1 ln ∣ x − 1∣ − 2 1 ln ∣ x + 1∣ + C = 2 1 ln x + 1 x − 1 + C An improper integral involves either an infinite limit of integration or an infinite discontinuity In the interval.
Type 1: Infinite interval:
∫ a ∞ f ( x ) d x = lim b → ∞ ∫ a b f ( x ) d x \int_a^{\infty} f(x)\, dx = \lim_{b \to \infty} \int_a^b f(x)\, dx ∫ a ∞ f ( x ) d x = b → ∞ lim ∫ a b f ( x ) d x Type 2: Infinite discontinuity at a a a
∫ a b f ( x ) d x = lim t → a + ∫ t b f ( x ) d x \int_a^b f(x)\, dx = \lim_{t \to a^+} \int_t^b f(x)\, dx ∫ a b f ( x ) d x = t → a + lim ∫ t b f ( x ) d x Example
Determine whether ∫ 1 ∞ 1 x p d x \displaystyle\int_1^{\infty} \frac{1}{x^p}\, dx ∫ 1 ∞ x p 1 d x p > 0 p \gt 0 p > 0
If p ≠ 1 p \ne 1 p = 1
= lim b → ∞ [ x 1 − p 1 − p ] 1 b = lim b → ∞ b 1 − p − 1 1 − p = \lim_{b \to \infty} \left[\frac{x^{1-p}}{1-p}\right]_1^b = \lim_{b \to \infty} \frac{b^{1-p} - 1}{1-p} = b → ∞ lim [ 1 − p x 1 − p ] 1 b = b → ∞ lim 1 − p b 1 − p − 1 This converges to 1 p − 1 \frac{1}{p-1} p − 1 1 p > 1 p \gt 1 p > 1 p < 1 p \lt 1 p < 1
If p = 1 p = 1 p = 1
\lim_{b \to \infty} [\ln x]_1^b = \lim_{b \to \infty} \ln b = \infty \quad \mathrm{(diverges) Example
Evaluate ∫ 0 ∞ x e − x 2 d x \displaystyle\int_0^{\infty} xe^{-x^2}\, dx ∫ 0 ∞ x e − x 2 d x
Let u = x^2$$du = 2x\, dx :
∫ 0 ∞ x e − x 2 d x = 1 2 ∫ 0 ∞ e − u d u = 1 2 [ − e − u ] 0 ∞ = 1 2 ( 0 + 1 ) = 1 2 \int_0^{\infty} xe^{-x^2}\, dx = \frac{1}{2}\int_0^{\infty} e^{-u}\, du = \frac{1}{2}\left[-e^{-u}\right]_0^{\infty} = \frac{1}{2}(0 + 1) = \frac{1}{2} ∫ 0 ∞ x e − x 2 d x = 2 1 ∫ 0 ∞ e − u d u = 2 1 [ − e − u ] 0 ∞ = 2 1 ( 0 + 1 ) = 2 1 The Gaussian integral. The integral ∫ 0 ∞ e − x 2 d x = π 2 \displaystyle\int_0^{\infty} e^{-x^2}\, dx = \frac{\sqrt{\pi}}{2} ∫ 0 ∞ e − x 2 d x = 2 π − ∞ -\infty − ∞ ∞ \infty ∞ π \sqrt{\pi} π
Note that ∫ 0 ∞ x e − x 2 d x = 1 2 \displaystyle\int_0^{\infty} xe^{-x^2}\, dx = \frac{1}{2} ∫ 0 ∞ x e − x 2 d x = 2 1 u u u ∫ 0 ∞ e − x 2 d x = π 2 \displaystyle\int_0^{\infty} e^{-x^2}\, dx = \frac{\sqrt{\pi}}{2} ∫ 0 ∞ e − x 2 d x = 2 π
Example
Evaluate ∫ 0 ∞ e − x d x \displaystyle\int_0^{\infty} e^{-x}\, dx ∫ 0 ∞ e − x d x
lim b → ∞ [ − e − x ] 0 b = lim b → ∞ ( − e − b + 1 ) = 1 \lim_{b \to \infty} \left[-e^{-x}\right]_0^b = \lim_{b \to \infty}(-e^{-b} + 1) = 1 b → ∞ lim [ − e − x ] 0 b = b → ∞ lim ( − e − b + 1 ) = 1 The integral converges to 1 1 1
The area between y = f ( x ) y = f(x) y = f ( x ) y = g ( x ) y = g(x) y = g ( x ) x = a x = a x = a x = b x = b x = b f ( x ) ≥ g ( x ) f(x) \geq g(x) f ( x ) ≥ g ( x )
A = ∫ a b [ f ( x ) − g ( x ) ] d x A = \int_a^b [f(x) - g(x)]\, dx A = ∫ a b [ f ( x ) − g ( x )] d x When to split. If f f f g g g
Example
Find the area between y = x 2 y = x^2 y = x 2 y = 2 x y = 2x y = 2 x
Find intersections: x 2 = 2 x x^2 = 2x x 2 = 2 x x 2 − 2 x = 0 x^2 - 2x = 0 x 2 − 2 x = 0 x = 0 x = 0 x = 0 x = 2 x = 2 x = 2
Between x = 0 x = 0 x = 0 x = 2$$2x \ge x^2 .
A = ∫ 0 2 ( 2 x − x 2 ) d x = [ x 2 − x 3 3 ] 0 2 = 4 − 8 3 = 4 3 A = \int_0^2 (2x - x^2)\, dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3} A = ∫ 0 2 ( 2 x − x 2 ) d x = [ x 2 − 3 x 3 ] 0 2 = 4 − 3 8 = 3 4 Disk method (rotating about the x x x
V = π ∫ a b [ f ( x ) ] 2 d x V = \pi \int_a^b [f(x)]^2\, dx V = π ∫ a b [ f ( x ) ] 2 d x Washer method (rotating region between f ( x ) f(x) f ( x ) g ( x ) g(x) g ( x ) x x x
V = π ∫ a b ( [ f ( x ) ] 2 − [ g ( x ) ] 2 ) d x V = \pi \int_a^b \left([f(x)]^2 - [g(x)]^2\right)\, dx V = π ∫ a b ( [ f ( x ) ] 2 − [ g ( x ) ] 2 ) d x Shell method (rotating about the y y y
V = 2 π ∫ a b x ⋅ f ( x ) d x V = 2\pi \int_a^b x \cdot f(x)\, dx V = 2 π ∫ a b x ⋅ f ( x ) d x When to use which method. Use the disk/washer method when integrating perpendicular to the axis Of rotation. Use the shell method when integrating parallel to the axis of rotation.
Example
Find the volume of the solid obtained by rotating y = \sqrt{x}$$y = 0$$x = 4 about the x x x
Using the disk method:
V = π ∫ 0 4 ( x ) 2 d x = π ∫ 0 4 x d x = π [ x 2 2 ] 0 4 = 8 π V = \pi \int_0^4 (\sqrt{x})^2\, dx = \pi \int_0^4 x\, dx = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi V = π ∫ 0 4 ( x ) 2 d x = π ∫ 0 4 x d x = π [ 2 x 2 ] 0 4 = 8 π Example
Find the volume of the solid obtained by rotating the region bounded by y = x 2 y = x^2 y = x 2 y = x y = x y = x y y y
The curves intersect at x = 0 x = 0 x = 0 x = 1 x = 1 x = 1
V = 2 π ∫ 0 1 x ( x − x 2 ) d x = 2 π ∫ 0 1 ( x 2 − x 3 ) d x = 2 π [ x 3 3 − x 4 4 ] 0 1 = 2 π ( 1 3 − 1 4 ) = 2 π ⋅ 1 12 = π 6 V = 2\pi \int_0^1 x(x - x^2)\, dx = 2\pi \int_0^1 (x^2 - x^3)\, dx = 2\pi\left[\frac{x^3}{3} - \frac{x^4}{4}\right]_0^1 = 2\pi\left(\frac{1}{3} - \frac{1}{4}\right) = 2\pi \cdot \frac{1}{12} = \frac{\pi}{6} V = 2 π ∫ 0 1 x ( x − x 2 ) d x = 2 π ∫ 0 1 ( x 2 − x 3 ) d x = 2 π [ 3 x 3 − 4 x 4 ] 0 1 = 2 π ( 3 1 − 4 1 ) = 2 π ⋅ 12 1 = 6 π Example
Find the volume of the solid obtained by rotating the region bounded by y = e^{-x}$$y = 0 x = 0$$x = 1 about the x x x
Using the disk method:
V = π ∫ 0 1 ( e − x ) 2 d x = π ∫ 0 1 e − 2 x d x = π [ − e − 2 x 2 ] 0 1 = π ( − e − 2 2 + 1 2 ) = π 2 ( 1 − 1 e 2 ) V = \pi \int_0^1 (e^{-x})^2\, dx = \pi \int_0^1 e^{-2x}\, dx = \pi\left[-\frac{e^{-2x}}{2}\right]_0^1 = \pi\left(-\frac{e^{-2}}{2} + \frac{1}{2}\right) = \frac{\pi}{2}\!\left(1 - \frac{1}{e^2}\right) V = π ∫ 0 1 ( e − x ) 2 d x = π ∫ 0 1 e − 2 x d x = π [ − 2 e − 2 x ] 0 1 = π ( − 2 e − 2 + 2 1 ) = 2 π ( 1 − e 2 1 ) The average value of f f f [ a , b ] [a, b] [ a , b ]
F_{\mathrm{avg} = \frac{1}{b - a}\int_a^b f(x)\, dx By the MVT for integrals, there exists c ∈ [ a , b ] c \in [a, b] c ∈ [ a , b ] f(c) = f_{\mathrm{avg} .
Example
Find the average value of f ( x ) = sin x f(x) = \sin x f ( x ) = sin x [ 0 , π ] [0, \pi] [ 0 , π ]
F_{\mathrm{avg} = \frac{1}{\pi}\int_0^{\pi} \sin x\, dx = \frac{1}{\pi}[-\cos x]_0^{\pi} = \frac{1}{\pi}(1 - (-1)) = \frac{2}{\pi} The arc length of y = f ( x ) y = f(x) y = f ( x ) x = a x = a x = a x = b x = b x = b
L = ∫ a b 1 + [ f ′ ( x ) ] 2 d x L = \int_a^b \sqrt{1 + [f'(x)]^2}\, dx L = ∫ a b 1 + [ f ′ ( x ) ] 2 d x Example
Find the arc length of y = 2 3 x 3 / 2 y = \frac{2}{3}x^{3/2} y = 3 2 x 3/2 x = 0 x = 0 x = 0 x = 3 x = 3 x = 3
f ′ ( x ) = x 1 / 2 = x f'(x) = x^{1/2} = \sqrt{x} f ′ ( x ) = x 1/2 = x
L = ∫ 0 3 1 + x d x L = \int_0^3 \sqrt{1 + x}\, dx L = ∫ 0 3 1 + x d x Let u = 1 + x$$du = dx :
L = ∫ 1 4 u d u = [ 2 u 3 / 2 3 ] 1 4 = 2 3 ( 8 − 1 ) = 14 3 L = \int_1^4 \sqrt{u}\, du = \left[\frac{2u^{3/2}}{3}\right]_1^4 = \frac{2}{3}(8 - 1) = \frac{14}{3} L = ∫ 1 4 u d u = [ 3 2 u 3/2 ] 1 4 = 3 2 ( 8 − 1 ) = 3 14 Forgetting the constant of integration in indefinite integrals. Without it, the answer is incomplete.Incorrect u u u When using u u u Forgetting to change d x dx d x d u du d u when performing u u u Applying FTC Part 2 to discontinuous functions. f f f [ a , b ] [a, b] [ a , b ] Confusing net area with total area. Use absolute values for total area.Incorrectly choosing u u u Use LIATE: prioritize Logarithmic over Inverse trig over Algebraic over Trig over Exponential.Sign errors with FTC Part 1 chain rule. The derivative of ∫ a g ( x ) f ( t ) d t \int_a^{g(x)} f(t)\, dt ∫ a g ( x ) f ( t ) d t f ( g ( x ) ) ⋅ g ′ ( x ) f(g(x)) \cdot g'(x) f ( g ( x )) ⋅ g ′ ( x ) f ( g ( x ) ) f(g(x)) f ( g ( x )) Forgetting to split improper integrals at the discontinuity when a singularity is in the interval.Choosing the disk vs. Washer vs. Shell method incorrectly. Disk: rotate around x x x y y y Dropping the absolute value in ln ∣ x ∣ \ln|x| ln ∣ x ∣ The antiderivative of 1 x \frac{1}{x} x 1 ln ∣ x ∣ + C \ln|x| + C ln ∣ x ∣ + C ln x + C \ln x + C ln x + C x < 0 x \lt 0 x < 0 ln ( − x ) + C \ln(-x) + C ln ( − x ) + C Confusing the Gaussian integrals. ∫ 0 ∞ x e − x 2 d x = 1 2 \displaystyle\int_0^{\infty} xe^{-x^2}\, dx = \frac{1}{2} ∫ 0 ∞ x e − x 2 d x = 2 1 u u u ∫ 0 ∞ e − x 2 d x = π 2 \displaystyle\int_0^{\infty} e^{-x^2}\, dx = \frac{\sqrt{\pi}}{2} ∫ 0 ∞ e − x 2 d x = 2 π Evaluate ∫ x x 2 + 1 d x \displaystyle\int \frac{x}{x^2 + 1}\, dx ∫ x 2 + 1 x d x
Evaluate ∫ 0 π / 2 sin 2 x d x \displaystyle\int_0^{\pi/2} \sin^2 x\, dx ∫ 0 π /2 sin 2 x d x sin 2 x = 1 − cos 2 x 2 \sin^2 x = \frac{1 - \cos 2x}{2} sin 2 x = 2 1 − c o s 2 x
Use integration by parts to evaluate ∫ x 2 e x d x \displaystyle\int x^2 e^x\, dx ∫ x 2 e x d x
Find the area between y = x 2 y = x^2 y = x 2 y = 2 x y = 2x y = 2 x
Find the volume of the solid obtained by rotating the region bounded by y = x^2$$y = 0 x = 1 x = 1 x = 1 y y y
Determine whether ∫ 0 ∞ e − x d x \displaystyle\int_0^{\infty} e^{-x}\, dx ∫ 0 ∞ e − x d x
Find the average value of f ( x ) = sin x f(x) = \sin x f ( x ) = sin x [ 0 , π ] [0, \pi] [ 0 , π ]
Given g ( x ) = ∫ 0 x 3 cos ( t 2 ) d t \displaystyle g(x) = \int_0^{x^3} \cos(t^2)\, dt g ( x ) = ∫ 0 x 3 cos ( t 2 ) d t g ′ ( x ) g'(x) g ′ ( x )
Evaluate ∫ 0 1 2 x 1 + x 2 d x \displaystyle\int_0^1 \frac{2x}{\sqrt{1 + x^2}}\, dx ∫ 0 1 1 + x 2 2 x d x
Find the arc length of y = x 2 8 + 2 y = \frac{x^2}{8} + 2 y = 8 x 2 + 2 x = 0 x = 0 x = 0 x = 4 x = 4 x = 4
Evaluate ∫ 1 e ln x x d x \displaystyle\int_1^e \frac{\ln x}{x}\, dx ∫ 1 e x ln x d x
Find the volume when the region bounded by y = e^{-x}$$y = 0$$x = 0$$x = 1 is rotated about the x x x
Evaluate ∫ 0 1 1 1 + x 2 d x \displaystyle\int_0^1 \frac{1}{1 + x^2}\, dx ∫ 0 1 1 + x 2 1 d x
Use the substitution x = tan θ x = \tan\theta x = tan θ ∫ 1 1 + x 2 d x \displaystyle\int \frac{1}{1 + x^2}\, dx ∫ 1 + x 2 1 d x
Find the area between the curves y = sin x y = \sin x y = sin x y = cos x y = \cos x y = cos x 0 ≤ x ≤ π / 2 0 \le x \le \pi/2 0 ≤ x ≤ π /2
Evaluate ∫ 0 1 x 2 e − x d x \displaystyle\int_0^1 x^2 e^{-x}\, dx ∫ 0 1 x 2 e − x d x
Question 1: Integration by parts (tabular method) Evaluate ∫ x 3 e x d x \displaystyle\int x^3 e^x \, dx ∫ x 3 e x d x
Answer Differentiate x 3 x^3 x 3 e x e^x e x + , − , + , − +,-,+,- + , − , + , −
Signs x 3 x^3 x 3 e x e^x e x + x 3 x^3 x 3 e x e^x e x - 3 x 2 3x^2 3 x 2 e x e^x e x + 6 x 6x 6 x e x e^x e x - 6 6 6 e x e^x e x + 0 0 0 e x e^x e x
Result: x 3 e x − 3 x 2 e x + 6 x e x − 6 e x + C = e x ( x 3 − 3 x 2 + 6 x − 6 ) + C x^3 e^x - 3x^2 e^x + 6xe^x - 6e^x + C = e^x(x^3 - 3x^2 + 6x - 6) + C x 3 e x − 3 x 2 e x + 6 x e x − 6 e x + C = e x ( x 3 − 3 x 2 + 6 x − 6 ) + C
Question 2: Trigonometric substitution Evaluate ∫ x 2 9 − x 2 d x \displaystyle\int \frac{x^2}{\sqrt{9 - x^2}} \, dx ∫ 9 − x 2 x 2 d x
Answer Let x = 3\sin\theta$$dx = 3\cos\theta \, d\theta$$\sqrt{9 - x^2} = 3\cos\theta .
= ∫ 9 sin 2 θ 3 cos θ ⋅ 3 cos θ d θ = 9 ∫ sin 2 θ d θ = 9 ∫ 1 − cos 2 θ 2 d θ = \int \frac{9\sin^2\theta}{3\cos\theta} \cdot 3\cos\theta \, d\theta = 9\int \sin^2\theta \, d\theta = 9\int \frac{1 - \cos 2\theta}{2} \, d\theta = ∫ 3 c o s θ 9 s i n 2 θ ⋅ 3 cos θ d θ = 9 ∫ sin 2 θ d θ = 9 ∫ 2 1 − c o s 2 θ d θ
= 9 2 ( θ − sin 2 θ 2 ) + C = 9 2 θ − 9 4 sin 2 θ + C = \frac{9}{2}\left(\theta - \frac{\sin 2\theta}{2}\right) + C = \frac{9}{2}\theta - \frac{9}{4}\sin 2\theta + C = 2 9 ( θ − 2 s i n 2 θ ) + C = 2 9 θ − 4 9 sin 2 θ + C
Since θ = arcsin ( x / 3 ) \theta = \arcsin(x/3) θ = arcsin ( x /3 ) sin 2 θ = 2 sin θ cos θ = 2 x 9 − x 2 9 \sin 2\theta = 2\sin\theta\cos\theta = \frac{2x\sqrt{9-x^2}}{9} sin 2 θ = 2 sin θ cos θ = 9 2 x 9 − x 2
= 9 2 arcsin ( x 3 ) − x 9 − x 2 2 + C = \frac{9}{2}\arcsin\!\left(\frac{x}{3}\right) - \frac{x\sqrt{9-x^2}}{2} + C = 2 9 arcsin ( 3 x ) − 2 x 9 − x 2 + C
Question 3: Area between curves Find the area enclosed by the curves y = x 2 y = x^2 y = x 2 y = 2 x + 3 y = 2x + 3 y = 2 x + 3
Answer Intersection: x^2 = 2x + 3$$x^2 - 2x - 3 = 0$$(x-3)(x+1) = 0 . x = − 1 x = -1 x = − 1 x = 3 x = 3 x = 3
On [-1, 3]$$2x + 3 \ge x^2 .
Area = ∫ − 1 3 [ ( 2 x + 3 ) − x 2 ] d x = [ x 2 + 3 x − x 3 3 ] − 1 3 = \int_{-1}^{3} [(2x + 3) - x^2] \, dx = \left[x^2 + 3x - \frac{x^3}{3}\right]_{-1}^{3} = ∫ − 1 3 [( 2 x + 3 ) − x 2 ] d x = [ x 2 + 3 x − 3 x 3 ] − 1 3
= ( 9 + 9 − 9 ) − ( 1 − 3 + 1 / 3 ) = 9 − ( − 5 / 3 ) = 9 + 5 / 3 = 32 / 3 = (9 + 9 - 9) - (1 - 3 + 1/3) = 9 - (-5/3) = 9 + 5/3 = 32/3 = ( 9 + 9 − 9 ) − ( 1 − 3 + 1/3 ) = 9 − ( − 5/3 ) = 9 + 5/3 = 32/3
Question 4: Improper integral Determine whether ∫ 1 ∞ 1 x p d x \displaystyle\int_1^{\infty} \frac{1}{x^p} \, dx ∫ 1 ∞ x p 1 d x
Answer ∫ 1 ∞ 1 x p d x = lim b → ∞ ∫ 1 b x − p d x \displaystyle\int_1^{\infty} \frac{1}{x^p} \, dx = \lim_{b \to \infty} \int_1^b x^{-p} \, dx ∫ 1 ∞ x p 1 d x = b → ∞ lim ∫ 1 b x − p d x
For p ≠ 1 p \ne 1 p = 1 = lim b → ∞ [ x 1 − p 1 − p ] 1 b = lim b → ∞ b 1 − p − 1 1 − p = \lim_{b \to \infty} \left[\frac{x^{1-p}}{1-p}\right]_1^b = \lim_{b \to \infty} \frac{b^{1-p} - 1}{1-p} = lim b → ∞ [ 1 − p x 1 − p ] 1 b = lim b → ∞ 1 − p b 1 − p − 1
If p > 1 p \gt 1 p > 1 1 − p < 0 1 - p \lt 0 1 − p < 0 b 1 − p → 0 b^{1-p} \to 0 b 1 − p → 0 1 p − 1 \frac{1}{p - 1} p − 1 1 If p < 1 p \lt 1 p < 1 1 − p > 0 1 - p \gt 0 1 − p > 0 b 1 − p → ∞ b^{1-p} \to \infty b 1 − p → ∞ For p = 1 p = 1 p = 1 ∫ 1 b 1 x d x = ln b → ∞ \int_1^b \frac{1}{x} \, dx = \ln b \to \infty ∫ 1 b x 1 d x = ln b → ∞
The integral converges if and only if p > 1 p \gt 1 p > 1 1 p − 1 \frac{1}{p-1} p − 1 1
Question 5: Volume of revolution Find the volume obtained by rotating the region bounded by y = \sqrt{x}$$y = 0 And x = 4 x = 4 x = 4
Answer Using the disk method:
V = π ∫ 0 4 ( x ) 2 d x = π ∫ 0 4 x d x = π [ x 2 2 ] 0 4 = π ( 8 ) = 8 π V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx = \pi \left[\frac{x^2}{2}\right]_0^4 = \pi(8) = 8\pi V = π ∫ 0 4 ( x ) 2 d x = π ∫ 0 4 x d x = π [ 2 x 2 ] 0 4 = π ( 8 ) = 8 π
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This topic covers the mathematical techniques and concepts related to integrals, including key theorems, methods, and problem-solving approaches.
Key concepts include:
differentiation from first principles product, quotient, and chain rules integration techniques (by parts, substitution) differential equations applications to kinematics Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.
Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.