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Integration

In mathematics for economists, we often need to move from marginal values (slopes) back to total values.

Reversing the process of differentiation to find the original function from its derivative is called integration or reverse differentiation.

The original function F(x)F(x) is called the integral or antiderivative of f′(x)f'(x).

Symbolically:

∫f(x)dx=F(x)+c\int f(x)dx = F(x) + c
(1)

This chapter covers:

  1. Basic rules of integration

  2. Integration by substitution

  3. Integration by parts

  4. Definite integrals and the Fundamental Theorem of Calculus

  5. Economic applications

Rules of Integration

  1. The integral of a constant kk is:

∫kdx=kx+c\int kdx = kx + c

  1. The integral of 1, written simply as dxdx, not 1dx1 dx, is:

∫dx=x+c\int dx = x + c

  1. The integral of a power function xnx^n, where n≠−1n \neq -1, is given by the power rule:

∫xndx=1n+1xn+1+c\int x^n dx = \frac{1}{n+1}x^{n+1} + c

  1. The integral of x−1x^{-1} (or 1/x1/x) is:

∫x−1dx=ln⁡x+c(x>0)\int x^{-1} dx = \ln x + c \quad (x > 0)

The condition x>0x > 0 is added because only positive numbers have logarithms. For negative numbers,

∫x−1dx=ln⁡∣x∣+c(x≠0)\int x^{-1} dx = \ln|x| + c \quad (x \neq 0)

A little thought informs us that:

∫f′(x)f(x) dx=ln⁡f(x)+Cfor f(x)>0=ln⁡∣f(x)∣+Cfor f(x)≠0\begin{align} \int \frac{f'(x)}{f(x)} \, dx &= \ln f(x) + C \quad \text{for } f(x) > 0 \\ &= \ln |f(x)| + C \quad \text{for } f(x) \neq 0 \end{align}

  1. The integral of a exponential function is

∫akxdx=akxkln⁡x+c\int a^{kx} dx = \frac{a^{kx}}{k \ln x} + c

  1. The integral of a natural exponential function is

∫ekxdx=ekxk+c\int e^{kx} dx = \frac{e^{kx}}{k} + c

From our knowledge of differentiation of exponential functions, it goes without saying that

∫exdx=ex+c\int e^x dx = e^x + c

And with a little thought, we have

∫f′(x)ef(x)dx=ef(x)+c\int f'(x)e^{f(x)} dx = e^{f(x)} + c

  1. The integral of a constant times a function equals the constant times the integral of the function:

∫kf(x)dx=k∫f(x)dx\int kf(x) dx = k \int f(x) dx

  1. The integral of the sum or difference of two or more functions equals the sum or difference of their integrals:

∫[f(x)±g(x)]dx=∫f(x)dx±∫g(x)dx\int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx

  1. The integral of the negative of a function equals the negative of the integral of that function:

∫−f(x)dx=−∫f(x)dx\int -f(x) dx = -\int f(x) dx
Try these ^^

Determine the following integrals. You should also check your answers by finding the derivative of the integral that you find.

i. ∫−12xdx\int -\frac{1}{2}x dx

ii. ∫4x3dx\int 4x^3 dx

iii. ∫4x−2dx\int 4x^{-2} dx

iv. ∫x−5/2dx\int x^{-5/2} dx

v. ∫5x−1dx\int 5x^{-1} dx

vi. ∫13xdx\int \frac{1}{3x} dx

vii. ∫dxx3\int \frac{dx}{\sqrt[3]{x}}

viii. ∫24xdx\int 2^{4x} dx

ix. ∫8xdx\int 8^x dx

x. ∫e5xdx\int e^{5x} dx

xi. ∫16e−4xdx\int 16e^{-4x} dx

xii. ∫(6e3x−8e−2x)dx\int (6e^{3x} - 8e^{-2x}) dx

  1. Find the integral for y=∫(x1/2+3x−1/2)dxy = \int(x^{1/2} + 3x^{-1/2})dx, given the initial condition y=0y = 0 when x=0x = 0.

    answer: y=23x3/2+6x1/2y = \frac{2}{3}x^{3/2} + 6x^{1/2}

  2. Find the integral for y=∫(2x5−3x−1/4)dxy = \int(2x^5 - 3x^{-1/4})dx, given the initial condition y=6y = 6 when x=0x = 0.

    answer: y=13x6−4x3/4+6y = \frac{1}{3}x^6 - 4x^{3/4} + 6

  3. Find the integral for y=∫(10x4−3)dxy = \int(10x^4 - 3)dx, given the boundary condition y=21y = 21 when x=1x = 1.

    answer: y=2x5−3x+22y = 2x^5 - 3x + 22

Solutions to Exercises

i. ∫−12x dx\int -\frac{1}{2}x \, dx

  • Step: Apply the power rule: ∫xndx=1n+1xn+1+c\int x^n dx = \frac{1}{n+1}x^{n+1} + c.

  • Solution: −12⋅x22+c=−14x2+c-\frac{1}{2} \cdot \frac{x^2}{2} + c = -\frac{1}{4}x^2 + c

  • Check: ddx(−14x2+c)=−14(2x)=−12x\frac{d}{dx}(-\frac{1}{4}x^2 + c) = -\frac{1}{4}(2x) = -\frac{1}{2}x

ii. ∫4x3 dx\int 4x^3 \, dx

  • Step: Factor out the constant and apply the power rule.

  • Solution: 4⋅x44+c=x4+c4 \cdot \frac{x^4}{4} + c = x^4 + c

  • Check: ddx(x4+c)=4x3\frac{d}{dx}(x^4 + c) = 4x^3

iii. ∫4x−2 dx\int 4x^{-2} \, dx

  • Step: Apply the power rule for n=−2n = -2.

  • Solution: 4⋅x−1−1+c=−4x−1+c=−4x+c4 \cdot \frac{x^{-1}}{-1} + c = -4x^{-1} + c = -\frac{4}{x} + c

  • Check: ddx(−4x−1+c)=−4(−1)x−2=4x−2\frac{d}{dx}(-4x^{-1} + c) = -4(-1)x^{-2} = 4x^{-2}

iv. ∫x−5/2 dx\int x^{-5/2} \, dx

  • Step: Apply the power rule: n+1=−5/2+2/2=−3/2n+1 = -5/2 + 2/2 = -3/2.

  • Solution: x−3/2−3/2+c=−23x−3/2+c\frac{x^{-3/2}}{-3/2} + c = -\frac{2}{3}x^{-3/2} + c

  • Check: ddx(−23x−3/2+c)=−23(−32)x−5/2=x−5/2\frac{d}{dx}(-\frac{2}{3}x^{-3/2} + c) = -\frac{2}{3}(-\frac{3}{2})x^{-5/2} = x^{-5/2}

v. ∫5x−1 dx\int 5x^{-1} \, dx

  • Step: Apply the log rule: ∫x−1dx=ln⁡∣x∣+c\int x^{-1} dx = \ln|x| + c.

  • Solution: 5ln⁡∣x∣+c5 \ln|x| + c

  • Check: ddx(5ln⁡∣x∣+c)=5⋅1x=5x−1\frac{d}{dx}(5 \ln|x| + c) = 5 \cdot \frac{1}{x} = 5x^{-1}

vi. ∫13x dx\int \frac{1}{3x} \, dx

  • Step: Factor out 1/31/3 and apply the log rule.

  • Solution: 13∫1x dx=13ln⁡∣x∣+c\frac{1}{3} \int \frac{1}{x} \, dx = \frac{1}{3} \ln|x| + c

  • Check: ddx(13ln⁡∣x∣+c)=13⋅1x=13x\frac{d}{dx}(\frac{1}{3} \ln|x| + c) = \frac{1}{3} \cdot \frac{1}{x} = \frac{1}{3x}

vii. ∫dxx3\int \frac{dx}{\sqrt[3]{x}}

  • Step: Rewrite as a power: x−1/3x^{-1/3}, then apply the power rule (n+1=2/3n+1 = 2/3).

  • Solution: ∫x−1/3 dx=x2/32/3+c=32x2/3+c\int x^{-1/3} \, dx = \frac{x^{2/3}}{2/3} + c = \frac{3}{2}x^{2/3} + c

  • Check: ddx(32x2/3+c)=32(23)x−1/3=x−1/3=1x3\frac{d}{dx}(\frac{3}{2}x^{2/3} + c) = \frac{3}{2}(\frac{2}{3})x^{-1/3} = x^{-1/3} = \frac{1}{\sqrt[3]{x}}

viii. ∫24x dx\int 2^{4x} \, dx

  • Step: Apply the general exponential rule: ∫akxdx=akxkln⁡a+c\int a^{kx} dx = \frac{a^{kx}}{k \ln a} + c.

  • Solution: 24x4ln⁡2+c\frac{2^{4x}}{4 \ln 2} + c

  • Check: ddx(24x4ln⁡2+c)=14ln⁡2⋅24x⋅ln⁡2⋅4=24x\frac{d}{dx}(\frac{2^{4x}}{4 \ln 2} + c) = \frac{1}{4 \ln 2} \cdot 2^{4x} \cdot \ln 2 \cdot 4 = 2^{4x}

ix. ∫8x dx\int 8^x \, dx

  • Step: Apply the exponential rule where k=1k=1.

  • Solution: 8xln⁡8+c\frac{8^x}{\ln 8} + c

  • Check: ddx(8xln⁡8+c)=1ln⁡8⋅8x⋅ln⁡8=8x\frac{d}{dx}(\frac{8^x}{\ln 8} + c) = \frac{1}{\ln 8} \cdot 8^x \cdot \ln 8 = 8^x

x. ∫e5x dx\int e^{5x} \, dx

  • Step: Apply the natural exponential rule: ∫ekxdx=ekxk+c\int e^{kx} dx = \frac{e^{kx}}{k} + c.

  • Solution: 15e5x+c\frac{1}{5}e^{5x} + c

  • Check: ddx(15e5x+c)=15⋅e5x⋅5=e5x\frac{d}{dx}(\frac{1}{5}e^{5x} + c) = \frac{1}{5} \cdot e^{5x} \cdot 5 = e^{5x}

xi. ∫16e−4x dx\int 16e^{-4x} \, dx

  • Step: Factor out 16 and apply the natural exponential rule.

  • Solution: 16⋅e−4x−4+c=−4e−4x+c16 \cdot \frac{e^{-4x}}{-4} + c = -4e^{-4x} + c

  • Check: ddx(−4e−4x+c)=−4⋅e−4x⋅(−4)=16e−4x\frac{d}{dx}(-4e^{-4x} + c) = -4 \cdot e^{-4x} \cdot (-4) = 16e^{-4x}

xii. ∫(6e3x−8e−2x) dx\int (6e^{3x} - 8e^{-2x}) \, dx

  • Step: Apply the sum/difference rule and integrate each term individually.

  • Solution: 6⋅e3x3−8⋅e−2x−2+c=2e3x+4e−2x+c6 \cdot \frac{e^{3x}}{3} - 8 \cdot \frac{e^{-2x}}{-2} + c = 2e^{3x} + 4e^{-2x} + c

  • Check: ddx(2e3x+4e−2x+c)=2(3)e3x+4(−2)e−2x=6e3x−8e−2x\frac{d}{dx}(2e^{3x} + 4e^{-2x} + c) = 2(3)e^{3x} + 4(-2)e^{-2x} = 6e^{3x} - 8e^{-2x}

Integration by Substitution

Integration by substitution is often useful to integrate more complex forms.

∫f(g(x))g′(x)dx=∫f(u)duwhere u=g(x)\int f(g(x))g'(x) dx = \int f(u) du \quad \text{where } u = g(x)
(2)

Example:

Say you want to find:

∫(x2+4)32xdx\int (x^2 + 4)^3 2x dx

The “trick” is to notice that if u=g(x)=x2+4u = g(x) = x^2 + 4, then g′(x)=2xg'(x) = 2x.

So we can write:

∫(x2+4)32xdx=∫u3dusince du=2xdx=u44+c=(x2+4)44+c\begin{align*} \int (x^2 + 4)^3 2x dx &= \int u^3 du \quad \text{since } du = 2x dx \\ &= \frac{u^4}{4} + c = \frac{(x^2 + 4)^4}{4} + c \end{align*}

A slightly more difficult example:

Find ∫xe−kx2dx\int xe^{-kx^2} dx.

Let u=−kx2u = -kx^2, then du=−2kxdxdu = -2kx dx.

Hence, xdx=−du2kx dx = -\frac{du}{2k}, implying:

∫xe−kx2dx=∫−12keudu=−12keu+c=−12ke−kx2+c\int xe^{-kx^2} dx = \int -\frac{1}{2k} e^u du = -\frac{1}{2k} e^u + c = -\frac{1}{2k} e^{-kx^2} + c

Generalized Rules from Substitution

Generalized Power Rule (n≠−1n \neq -1):

∫[f(x)]nf′(x)dx=[f(x)]n+1n+1+c\int [f(x)]^n f'(x) dx = \frac{[f(x)]^{n+1}}{n+1} + c

Example 1:

∫(x+1)2dx=(x+1)33+c\int (x+1)^2 dx = \frac{(x+1)^3}{3} + c

Example 2:

∫(2x+1)2dx=12∫(2x+1)22dx=(2x+1)36+c\int (2x+1)^2 dx = \frac{1}{2} \int (2x+1)^2 2 dx = \frac{(2x+1)^3}{6} + c

Generalized Log Rule:

∫f′(x)f(x)dx=ln⁡∣f(x)∣+c\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + c

Example 1:

∫2xdx=2ln⁡(x)+c(x>0)\int \frac{2}{x} dx = 2 \ln(x) + c \quad (x > 0)

Example 2:

∫x(4x2+1)dx=18∫8x(4x2+1)dx=18ln⁡(4x2+1)+c\int \frac{x}{(4x^2 + 1)} dx = \frac{1}{8} \int \frac{8x}{(4x^2 + 1)} dx = \frac{1}{8} \ln(4x^2 + 1) + c

Generalized Exponential Rule:

∫Aef(x)f′(x)dx=Aef(x)+c\int A e^{f(x)} f'(x) dx = A e^{f(x)} + c

Example:

∫xex2dx=12ex2+c\int x e^{x^2} dx = \frac{1}{2} e^{x^2} + c
Try these ^^

Determine the following integrals, using the substitution method. Check your answers.

i. ∫10x(x2+3)4dx\int 10x(x^2 + 3)^4 dx

ii. ∫x4(2x5−5)4dx\int x^4(2x^5 - 5)^4 dx

iii. ∫(x−9)7/4dx\int (x - 9)^{7/4} dx

iv. ∫(6x−11)−5dx\int (6x - 11)^{-5} dx

v. ∫x2(4x3+7)2dx\int \frac{x^2}{(4x^3 + 7)^2} dx

vi. ∫6x2+4x+10(x3+x2+5x)3dx\int \frac{6x^2 + 4x + 10}{(x^3 + x^2 + 5x)^3} dx

vii. ∫dx(9x−5)\int \frac{dx}{(9x - 5)}

viii. ∫3x2+24x3+8xdx\int \frac{3x^2 + 2}{4x^3 + 8x} dx

ix. ∫x3ex4dx\int x^3 e^{x^4} dx

x. ∫24xe3x2dx\int 24x e^{3x^2} dx

xi. ∫14e2x+7dx\int 14 e^{2x+7} dx

xii. ∫5xe5x2+3dx\int 5x e^{5x^2+3} dx

Solutions to Substitution Exercises

i. ∫10x(x2+3)4dx\int 10x(x^2 + 3)^4 dx

  • Step: Let u=x2+3u = x^2 + 3, then du=2x dxdu = 2x \, dx, so 10x dx=5 du10x \, dx = 5 \, du.

  • Solution: ∫5u4du=u5+c=(x2+3)5+c\int 5u^4 du = u^5 + c = (x^2 + 3)^5 + c

  • Check: ddx[(x2+3)5+c]=5(x2+3)4(2x)=10x(x2+3)4\frac{d}{dx}[(x^2 + 3)^5 + c] = 5(x^2 + 3)^4(2x) = 10x(x^2 + 3)^4

ii. ∫x4(2x5−5)4dx\int x^4(2x^5 - 5)^4 dx

  • Step: Let u=2x5−5u = 2x^5 - 5, then du=10x4 dxdu = 10x^4 \, dx, so x4 dx=110 dux^4 \, dx = \frac{1}{10} \, du.

  • Solution: ∫110u4du=150u5+c=150(2x5−5)5+c\int \frac{1}{10}u^4 du = \frac{1}{50}u^5 + c = \frac{1}{50}(2x^5 - 5)^5 + c

  • Check: ddx[150(2x5−5)5+c]=550(2x5−5)4(10x4)=x4(2x5−5)4\frac{d}{dx}[\frac{1}{50}(2x^5 - 5)^5 + c] = \frac{5}{50}(2x^5 - 5)^4(10x^4) = x^4(2x^5 - 5)^4

iii. ∫(x−9)7/4dx\int (x - 9)^{7/4} dx

  • Step: Let u=x−9u = x - 9, then du=dxdu = dx.

  • Solution: ∫u7/4du=u11/411/4+c=411(x−9)11/4+c\int u^{7/4} du = \frac{u^{11/4}}{11/4} + c = \frac{4}{11}(x - 9)^{11/4} + c

  • Check: ddx[411(x−9)11/4+c]=411⋅114(x−9)7/4=(x−9)7/4\frac{d}{dx}[\frac{4}{11}(x - 9)^{11/4} + c] = \frac{4}{11} \cdot \frac{11}{4}(x - 9)^{7/4} = (x - 9)^{7/4}

iv. ∫(6x−11)−5dx\int (6x - 11)^{-5} dx

  • Step: Let u=6x−11u = 6x - 11, then du=6 dxdu = 6 \, dx, so dx=16 dudx = \frac{1}{6} \, du.

  • Solution: ∫16u−5du=16(u−4−4)+c=−124(6x−11)−4+c\int \frac{1}{6}u^{-5} du = \frac{1}{6}(\frac{u^{-4}}{-4}) + c = -\frac{1}{24}(6x - 11)^{-4} + c

  • Check: ddx[−124(6x−11)−4+c]=−124(−4)(6x−11)−5(6)=(6x−11)−5\frac{d}{dx}[-\frac{1}{24}(6x - 11)^{-4} + c] = -\frac{1}{24}(-4)(6x - 11)^{-5}(6) = (6x - 11)^{-5}

v. ∫x2(4x3+7)2dx\int \frac{x^2}{(4x^3 + 7)^2} dx

  • Step: Let u=4x3+7u = 4x^3 + 7, then du=12x2 dxdu = 12x^2 \, dx, so x2 dx=112 dux^2 \, dx = \frac{1}{12} \, du.

  • Solution: ∫112u−2du=112(u−1−1)+c=−112(4x3+7)+c\int \frac{1}{12}u^{-2} du = \frac{1}{12}(\frac{u^{-1}}{-1}) + c = -\frac{1}{12(4x^3 + 7)} + c

  • Check: ddx[−112(4x3+7)−1+c]=−112(−1)(4x3+7)−2(12x2)=x2(4x3+7)2\frac{d}{dx}[-\frac{1}{12}(4x^3 + 7)^{-1} + c] = -\frac{1}{12}(-1)(4x^3 + 7)^{-2}(12x^2) = \frac{x^2}{(4x^3 + 7)^2}

vi. ∫6x2+4x+10(x3+x2+5x)3dx\int \frac{6x^2 + 4x + 10}{(x^3 + x^2 + 5x)^3} dx

  • Step: Let u=x3+x2+5xu = x^3 + x^2 + 5x, then du=(3x2+2x+5) dxdu = (3x^2 + 2x + 5) \, dx. Note the numerator is 2(3x2+2x+5)2(3x^2 + 2x + 5).

  • Solution: ∫2u−3du=2u−2−2+c=−(x3+x2+5x)−2+c=−1(x3+x2+5x)2+c\int 2u^{-3} du = \frac{2u^{-2}}{-2} + c = -(x^3 + x^2 + 5x)^{-2} + c = -\frac{1}{(x^3 + x^2 + 5x)^2} + c

  • Check: ddx[−(x3+x2+5x)−2+c]=−(−2)(x3+x2+5x)−3(3x2+2x+5)=6x2+4x+10(x3+x2+5x)3\frac{d}{dx}[-(x^3 + x^2 + 5x)^{-2} + c] = -(-2)(x^3 + x^2 + 5x)^{-3}(3x^2 + 2x + 5) = \frac{6x^2 + 4x + 10}{(x^3 + x^2 + 5x)^3}

vii. ∫dx(9x−5)\int \frac{dx}{(9x - 5)}

  • Step: Let u=9x−5u = 9x - 5, then du=9 dxdu = 9 \, dx, so dx=19 dudx = \frac{1}{9} \, du.

  • Solution: ∫191udu=19ln⁡∣9x−5∣+c\int \frac{1}{9} \frac{1}{u} du = \frac{1}{9} \ln|9x - 5| + c

  • Check: ddx[19ln⁡∣9x−5∣+c]=19⋅19x−5⋅9=19x−5\frac{d}{dx}[\frac{1}{9} \ln|9x - 5| + c] = \frac{1}{9} \cdot \frac{1}{9x - 5} \cdot 9 = \frac{1}{9x - 5}

viii. ∫3x2+24x3+8xdx\int \frac{3x^2 + 2}{4x^3 + 8x} dx

  • Step: Let u=4x3+8xu = 4x^3 + 8x, then du=(12x2+8) dx=4(3x2+2) dxdu = (12x^2 + 8) \, dx = 4(3x^2 + 2) \, dx.

  • Solution: ∫141udu=14ln⁡∣4x3+8x∣+c\int \frac{1}{4} \frac{1}{u} du = \frac{1}{4} \ln|4x^3 + 8x| + c

  • Check: ddx[14ln⁡∣4x3+8x∣+c]=14⋅12x2+84x3+8x=3x2+24x3+8x\frac{d}{dx}[\frac{1}{4} \ln|4x^3 + 8x| + c] = \frac{1}{4} \cdot \frac{12x^2 + 8}{4x^3 + 8x} = \frac{3x^2 + 2}{4x^3 + 8x}

ix. ∫x3ex4dx\int x^3 e^{x^4} dx

  • Step: Let u=x4u = x^4, then du=4x3 dxdu = 4x^3 \, dx, so x3 dx=14 dux^3 \, dx = \frac{1}{4} \, du.

  • Solution: ∫14eudu=14ex4+c\int \frac{1}{4} e^u du = \frac{1}{4} e^{x^4} + c

  • Check: ddx[14ex4+c]=14ex4(4x3)=x3ex4\frac{d}{dx}[\frac{1}{4} e^{x^4} + c] = \frac{1}{4} e^{x^4}(4x^3) = x^3 e^{x^4}

x. ∫24xe3x2dx\int 24x e^{3x^2} dx

  • Step: Let u=3x2u = 3x^2, then du=6x dxdu = 6x \, dx, so 24x dx=4 du24x \, dx = 4 \, du.

  • Solution: ∫4eudu=4e3x2+c\int 4e^u du = 4e^{3x^2} + c

  • Check: ddx[4e3x2+c]=4e3x2(6x)=24xe3x2\frac{d}{dx}[4e^{3x^2} + c] = 4e^{3x^2}(6x) = 24x e^{3x^2}

xi. ∫14e2x+7dx\int 14 e^{2x+7} dx

  • Step: Let u=2x+7u = 2x + 7, then du=2 dxdu = 2 \, dx, so 14 dx=7 du14 \, dx = 7 \, du.

  • Solution: ∫7eudu=7e2x+7+c\int 7e^u du = 7e^{2x+7} + c

  • Check: ddx[7e2x+7+c]=7e2x+7(2)=14e2x+7\frac{d}{dx}[7e^{2x+7} + c] = 7e^{2x+7}(2) = 14 e^{2x+7}

xii. ∫5xe5x2+3dx\int 5x e^{5x^2+3} dx

  • Step: Let u=5x2+3u = 5x^2 + 3, then du=10x dxdu = 10x \, dx, so 5x dx=12 du5x \, dx = \frac{1}{2} \, du.

  • Solution: ∫12eudu=12e5x2+3+c\int \frac{1}{2} e^u du = \frac{1}{2} e^{5x^2+3} + c

  • Check: ddx[12e5x2+3+c]=12e5x2+3(10x)=5xe5x2+3\frac{d}{dx}[\frac{1}{2} e^{5x^2+3} + c] = \frac{1}{2} e^{5x^2+3}(10x) = 5x e^{5x^2+3}

Integration by Parts

The product rule is:

ddx(f(x)g(x))=f′(x)g(x)+f(x)g′(x)\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)

Taking the indefinite integral of each side gives:

f(x)g(x)=∫f′(x)g(x)dx+∫f(x)g′(x)dxf(x)g(x) = \int f'(x)g(x) dx + \int f(x)g'(x) dx

Rearranging gives the formula for integration by parts:

∫f(x)g′(x)dx=f(x)g(x)−∫f′(x)g(x)dx\int f(x)g'(x) dx = f(x)g(x) - \int f'(x)g(x) dx
(3)

Example 1:

∫xexdx=xex−∫1⋅exdx=xex−ex+c\int x e^x dx = x e^x - \int 1 \cdot e^x dx = x e^x - e^x + c

Example 2:

∫xln⁡xdx=(ln⁡x)x22−∫1xx22dx=x22ln⁡x−∫x2dx=x22ln⁡x−x24+c\begin{align*} \int x \ln x dx &= (\ln x)\frac{x^2}{2} - \int \frac{1}{x} \frac{x^2}{2} dx \\ &= \frac{x^2}{2} \ln x - \int \frac{x}{2} dx \\ &= \frac{x^2}{2} \ln x - \frac{x^2}{4} + c \end{align*}
Try these ^^

  1. Use integration by parts to evaluate the following integrals. Check your answers.

i. ∫15x(x+4)3/2dx\int 15x(x+4)^{3/2} dx

ii. ∫2x(x−8)3dx\int \frac{2x}{(x - 8)^3} dx

iii. ∫5x(x−1)2dx\int \frac{5x}{(x - 1)^2} dx

iv. ∫6xex+7dx\int 6x e^{x+7} dx

v. ∫16xe−(x+9)dx\int 16x e^{-(x+9)} dx

vi. ∫x2e2xdx\int x^2 e^{2x} dx

Here are the step-by-step solutions for the Integration by Parts exercises.

These solutions use the formula ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du.

i. ∫15x(x+4)3/2dx\int 15x(x+4)^{3/2} dx

  • Step: Let u=15x⇒du=15 dxu = 15x \Rightarrow du = 15 \, dx. Let dv=(x+4)3/2dx⇒v=25(x+4)5/2dv = (x+4)^{3/2} dx \Rightarrow v = \frac{2}{5}(x+4)^{5/2}.

  • Solution: 15x⋅25(x+4)5/2−∫25(x+4)5/2⋅15 dx=6x(x+4)5/2−6∫(x+4)5/2dx15x \cdot \frac{2}{5}(x+4)^{5/2} - \int \frac{2}{5}(x+4)^{5/2} \cdot 15 \, dx = 6x(x+4)^{5/2} - 6 \int (x+4)^{5/2} dx

  • Final: 6x(x+4)5/2−127(x+4)7/2+c6x(x+4)^{5/2} - \frac{12}{7}(x+4)^{7/2} + c

  • Check: ddx\frac{d}{dx} yields 6(x+4)5/2+15x(x+4)3/2−6(x+4)5/2=15x(x+4)3/26(x+4)^{5/2} + 15x(x+4)^{3/2} - 6(x+4)^{5/2} = 15x(x+4)^{3/2}.

ii. ∫2x(x−8)3dx\int \frac{2x}{(x - 8)^3} dx

  • Step: Let u=2x⇒du=2 dxu = 2x \Rightarrow du = 2 \, dx. Let dv=(x−8)−3dx⇒v=−12(x−8)−2dv = (x-8)^{-3} dx \Rightarrow v = -\frac{1}{2}(x-8)^{-2}.

  • Solution: 2x[−12(x−8)−2]−∫−12(x−8)−2⋅2 dx=−x(x−8)−2+∫(x−8)−2dx2x [-\frac{1}{2}(x-8)^{-2}] - \int -\frac{1}{2}(x-8)^{-2} \cdot 2 \, dx = -x(x-8)^{-2} + \int (x-8)^{-2} dx

  • Final: −x(x−8)2−1x−8+c-\frac{x}{(x-8)^2} - \frac{1}{x-8} + c (Simplified: −2x−8(x−8)2+c-\frac{2x-8}{(x-8)^2} + c)

  • Check: Differentiating the simplified form returns 2x(x−8)3\frac{2x}{(x-8)^3}.

iii. ∫5x(x−1)2dx\int \frac{5x}{(x - 1)^2} dx

  • Step: Let u=5x⇒du=5 dxu = 5x \Rightarrow du = 5 \, dx. Let dv=(x−1)−2dx⇒v=−(x−1)−1dv = (x-1)^{-2} dx \Rightarrow v = -(x-1)^{-1}.

  • Solution: −5x(x−1)−1−∫−5(x−1)−1dx=−5xx−1+5ln⁡∣x−1∣+c-5x(x-1)^{-1} - \int -5(x-1)^{-1} dx = -\frac{5x}{x-1} + 5 \ln|x-1| + c

  • Check: ddx[−5xx−1+5ln⁡∣x−1∣]=−5(x−1)−(−5x)(x−1)2+5x−1=5(x−1)2+5(x−1)(x−1)2=5x(x−1)2\frac{d}{dx}[-\frac{5x}{x-1} + 5 \ln|x-1|] = \frac{-5(x-1) - (-5x)}{(x-1)^2} + \frac{5}{x-1} = \frac{5}{(x-1)^2} + \frac{5(x-1)}{(x-1)^2} = \frac{5x}{(x-1)^2}.

iv. ∫6xex+7dx\int 6x e^{x+7} dx

  • Step: Let u=6x⇒du=6 dxu = 6x \Rightarrow du = 6 \, dx. Let dv=ex+7dx⇒v=ex+7dv = e^{x+7} dx \Rightarrow v = e^{x+7}.

  • Solution: 6xex+7−∫6ex+7dx=6xex+7−6ex+7+c6x e^{x+7} - \int 6 e^{x+7} dx = 6x e^{x+7} - 6e^{x+7} + c

  • Final: 6ex+7(x−1)+c6e^{x+7}(x - 1) + c

  • Check: ddx[6xex+7−6ex+7]=6ex+7+6xex+7−6ex+7=6xex+7\frac{d}{dx}[6x e^{x+7} - 6e^{x+7}] = 6e^{x+7} + 6x e^{x+7} - 6e^{x+7} = 6x e^{x+7}.

v. ∫16xe−(x+9)dx\int 16x e^{-(x+9)} dx

  • Step: Let u=16x⇒du=16 dxu = 16x \Rightarrow du = 16 \, dx. Let dv=e−(x+9)dx⇒v=−e−(x+9)dv = e^{-(x+9)} dx \Rightarrow v = -e^{-(x+9)}.

  • Solution: −16xe−(x+9)−∫−16e−(x+9)dx=−16xe−(x+9)−16e−(x+9)+c-16x e^{-(x+9)} - \int -16 e^{-(x+9)} dx = -16x e^{-(x+9)} - 16e^{-(x+9)} + c

  • Final: −16e−(x+9)(x+1)+c-16e^{-(x+9)}(x + 1) + c

  • Check: ddx[−16xe−(x+9)−16e−(x+9)]=−16e−(x+9)+16xe−(x+9)+16e−(x+9)=16xe−(x+9)\frac{d}{dx}[-16x e^{-(x+9)} - 16e^{-(x+9)}] = -16e^{-(x+9)} + 16xe^{-(x+9)} + 16e^{-(x+9)} = 16x e^{-(x+9)}.

vi. ∫x2e2xdx\int x^2 e^{2x} dx

  • Step 1: Let u=x2⇒du=2x dxu = x^2 \Rightarrow du = 2x \, dx. Let dv=e2xdx⇒v=12e2xdv = e^{2x} dx \Rightarrow v = \frac{1}{2}e^{2x}.

  • Apply Parts 1: 12x2e2x−∫xe2xdx\frac{1}{2}x^2 e^{2x} - \int x e^{2x} dx

  • Step 2: Integrate ∫xe2xdx\int x e^{2x} dx by parts again. Let u=x⇒du=dxu = x \Rightarrow du = dx. Let dv=e2xdx⇒v=12e2xdv = e^{2x} dx \Rightarrow v = \frac{1}{2}e^{2x}.

  • Apply Parts 2: ∫xe2xdx=12xe2x−∫12e2xdx=12xe2x−14e2x\int x e^{2x} dx = \frac{1}{2}xe^{2x} - \int \frac{1}{2}e^{2x} dx = \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x}

  • Final Solution: 12x2e2x−(12xe2x−14e2x)+c=e2x(12x2−12x+14)+c\frac{1}{2}x^2 e^{2x} - (\frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x}) + c = e^{2x}(\frac{1}{2}x^2 - \frac{1}{2}x + \frac{1}{4}) + c

  • Check: Differentiating via product rule returns x2e2xx^2 e^{2x}.

Definite Integrals and The Fundamental Theorem of Calculus

The definite integral of f(x)f(x) over the interval aa to bb (a<x<ba < x < b) is written as:

∫abf(x)dx\int_{a}^{b} f(x) dx

The Fundamental Theorem of Calculus states that the numerical value of the definite integral of a continuous function f(x)f(x) over the interval from aa to bb is given by the indefinite integral F(x)+cF(x) + c evaluated at the upper limit bb, minus the same evaluated at the lower limit aa.

Since the constant cc is common to both, it is eliminated in subtraction:

∫abf(x)dx=F(x)∣ab=F(b)−F(a)\int_{a}^{b} f(x) dx = F(x) \Big|_{a}^{b} = F(b) - F(a)
(4)

Example:

∫13(4x3+6x)dx=(x4+3x2)∣13=[(34)+3(32)]−[(14)+3(12)]=108−4=104\begin{align*} \int_{1}^{3} (4x^3 + 6x) dx &= (x^4 + 3x^2) \Big|_{1}^{3} \\ &= [ (3^4) + 3(3^2) ] - [ (1^4) + 3(1^2) ] \\ &= 108 - 4 = 104 \end{align*}

This represents the area under the curve f(x)=4x3+6xf(x) = 4x^3 + 6x between x=1x=1 and x=3x=3.

Improper Integrals

A definite integral with infinity for either an upper or lower limit is known as an improper integral:

∫a∞f(x)dx\int_{a}^{\infty} f(x) dx

and

∫−∞bf(x)dx\int_{-\infty}^{b} f(x) dx

Properties of Definite Integrals

∫abf(x)dx=−∫baf(x)dx\int_{a}^{b} f(x) dx = -\int_{b}^{a} f(x) dx
∫aaf(x)dx=F(a)−F(a)=0\int_{a}^{a} f(x) dx = F(a) - F(a) = 0
∫acf(x)dx=∫abf(x)dx+∫bcf(x)dx\int_{a}^{c} f(x) dx = \int_{a}^{b} f(x) dx + \int_{b}^{c} f(x) dx
∫abf(x)dx±∫abg(x)dx=∫ab[f(x)±g(x)]dx\int_{a}^{b} f(x) dx \pm \int_{a}^{b} g(x) dx = \int_{a}^{b} [f(x) \pm g(x)] dx
∫abkf(x)dx=k∫abf(x)dx\int_{a}^{b} kf(x) dx = k \int_{a}^{b} f(x) dx

Economic Applications of Integral Calculus

Try these ^^

  1. The rate of net investment is I=40t3/5I = 40t^{3/5}, and the capital stock at t=0t = 0 is 75. Find the capital function KK.

    answer: K=25t8/5+75K = 25t^{8/5} + 75

  2. Marginal cost given by MC=dTC/dQ=25+30Q−9Q2\text{MC} = d\text{TC}/dQ = 25 + 30Q - 9Q^2. Fixed cost is 55. Find (i) total cost, (ii) average cost, and (iii) variable cost functions.

    answers: (i) TC=25Q+15Q2−3Q3+55\text{TC} = 25Q + 15Q^2 - 3Q^3 + 55 (ii) AC=TCQ=25+15Q−3Q2+55Q\text{AC} = \frac{\text{TC}}{Q} = 25 + 15Q - 3Q^2 + \frac{55}{Q} (iii) VC=TC−FC=25Q+15Q2−3Q3\text{VC} = \text{TC} - \text{FC} = 25Q + 15Q^2 - 3Q^3

  3. Marginal revenue is given by MR=dTR/dQ=60−2Q−2Q2\text{MR} = d\text{TR}/dQ = 60 - 2Q - 2Q^2. Find (i) the TR function and (ii) the demand function P=f(Q)P = f(Q).

    answers: (i) TR=60Q−Q2−23Q3\text{TR} = 60Q - Q^2 - \frac{2}{3}Q^3 (ii) P=AR=TR/Q=60−Q−23Q2P = \text{AR} = \text{TR}/Q = 60 - Q - \frac{2}{3}Q^2

  4. Given MC=16e0.4Q\text{MC} = 16e^{0.4Q} and FC=100\text{FC} = 100, find TC\text{TC}.

    answer: TC=40e0.4Q+60\text{TC} = 40e^{0.4Q} + 60

Step-by-step Solutions

1. Find Capital Function K(t)K(t) given I=40t3/5I = 40t^{3/5} and K(0)=75K(0) = 75

  • Step 1 (Integrate): K(t)=∫40t3/5dt=40(t8/58/5)+c=40(58)t8/5+c=25t8/5+cK(t) = \int 40t^{3/5} dt = 40 \left( \frac{t^{8/5}}{8/5} \right) + c = 40 \left( \frac{5}{8} \right) t^{8/5} + c = 25t^{8/5} + c.

  • Step 2 (Find cc): Use initial condition K(0)=75K(0) = 75. 25(0)8/5+c=75⇒c=7525(0)^{8/5} + c = 75 \Rightarrow c = 75.

  • Final: K=25t8/5+75K = 25t^{8/5} + 75.

2. Find Costs given MC=25+30Q−9Q2\text{MC} = 25 + 30Q - 9Q^2 and FC=55\text{FC} = 55

  • Total Cost (TC): ∫(25+30Q−9Q2)dQ=25Q+15Q2−3Q3+c\int (25 + 30Q - 9Q^2) dQ = 25Q + 15Q^2 - 3Q^3 + c. Since TC(0)=FC=55\text{TC}(0) = \text{FC} = 55, then c=55c = 55.

    • TC=25Q+15Q2−3Q3+55\text{TC} = 25Q + 15Q^2 - 3Q^3 + 55.

  • Average Cost (AC): AC=TCQ=25Q+15Q2−3Q3+55Q\text{AC} = \frac{\text{TC}}{Q} = \frac{25Q + 15Q^2 - 3Q^3 + 55}{Q}.

    • AC=25+15Q−3Q2+55Q\text{AC} = 25 + 15Q - 3Q^2 + \frac{55}{Q}.

  • Variable Cost (VC): VC=TC−FC\text{VC} = \text{TC} - \text{FC}.

    • VC=25Q+15Q2−3Q3\text{VC} = 25Q + 15Q^2 - 3Q^3.

3. Find TR and Demand given MR=60−2Q−2Q2\text{MR} = 60 - 2Q - 2Q^2

  • Total Revenue (TR): ∫(60−2Q−2Q2)dQ=60Q−Q2−23Q3+c\int (60 - 2Q - 2Q^2) dQ = 60Q - Q^2 - \frac{2}{3}Q^3 + c. Since TR=0\text{TR} = 0 when Q=0Q = 0, c=0c = 0.

    • TR=60Q−Q2−23Q3\text{TR} = 60Q - Q^2 - \frac{2}{3}Q^3.

  • Demand Function (P): P=Average Revenue=TRQP = \text{Average Revenue} = \frac{\text{TR}}{Q}.

    • P=60−Q−23Q2P = 60 - Q - \frac{2}{3}Q^2.

4. Find TC given MC=16e0.4Q\text{MC} = 16e^{0.4Q} and FC=100\text{FC} = 100

  • Step 1 (Integrate): TC=∫16e0.4QdQ=160.4e0.4Q+c=40e0.4Q+c\text{TC} = \int 16e^{0.4Q} dQ = \frac{16}{0.4}e^{0.4Q} + c = 40e^{0.4Q} + c.

  • Step 2 (Find cc): Set TC(0)=100\text{TC}(0) = 100. 40e0.4(0)+c=100⇒40(1)+c=100⇒c=6040e^{0.4(0)} + c = 100 \Rightarrow 40(1) + c = 100 \Rightarrow c = 60.

  • Final: TC=40e0.4Q+60\text{TC} = 40e^{0.4Q} + 60.

Math Econ
Constrained Optimization