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Chapter 7 · Integrals

Exercise 7.2 · Q17: ∫xex2 dx\int\frac{x}{e^{x^2}}\,dx

Exercise 7.2Integration by substitution · Section 7.3.1
The question

Integrate the function

xex2.\frac{x}{e^{x^2}}.
Exam answer

Let I=∫xex2 dxI = \displaystyle\int\frac{x}{e^{x^2}}\,dx⁠.

I=∫x e−x2 dxI = \int x\,e^{-x^2}\,dx

Put −x2=t-x^2 = t⁠. Then −2x dx=dt-2x\,dx = dt⁠.

I=−12∫et dt=−12et+CI = -\frac12\int e^t\,dt = -\frac12 e^t + C

∴ I=−12ex2+C\therefore\ I = -\dfrac{1}{2e^{x^2}} + C⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Dividing by ex2e^{x^2} is the same as multiplying by e−x2e^{-x^2}⁠. The integrand x e−x2x\,e^{-x^2} then contains the exponent −x2-x^2 and, up to a constant, its derivative −2x-2x⁠. So put t=−x2t = -x^2⁠; the integral becomes −12∫et dt-\frac12\int e^t\,dt⁠.

Given, to find, and the plan
Given
f(x)=xex2f(x) = \dfrac{x}{e^{x^2}}⁠. Since ex2>0e^{x^2} > 0 for every real xx⁠, ff is defined everywhere.
To find
∫xex2 dx\displaystyle\int \frac{x}{e^{x^2}}\,dx⁠.
Plan
(1) Rewrite as x e−x2x\,e^{-x^2}⁠. (2) Substitute t=−x2t = -x^2⁠. (3) Integrate ete^t⁠. (4) Return to xx and write the answer in the book's form. (5) Differentiate back.

Basics you need

1Negative exponents

1ea=e−a\frac{1}{e^a} = e^{-a} for every real aa⁠. Basics: powers

2Substitution

If t=g(x)t = g(x)⁠, then dt=g′(x) dxdt = g'(x)\,dx⁠; a constant multiple of g′(x) dxg'(x)\,dx in the integrand can be traded for dtdt⁠. Basics: substitution

3The exponential integral

∫et dt=et+C\int e^t\,dt = e^t + C (standard formula (xi), Section 7.2). Basics: integration undoes differentiation

Solution

  1. Rewrite the integrand

    xex2=x e−x2.\frac{x}{e^{x^2}} = x\,e^{-x^2}.
    In this form the integrand is visibly "something times e(exponent)e^{\text{(exponent)}}", and we can ask whether that something is the derivative of the exponent. The derivative of −x2-x^2 is −2x-2x⁠, which is −2-2 times the factor xx in front. So the substitution will work.
    1ea=e−a\frac{1}{e^{a}} = e^{-a}⁠, here with a=x2a = x^2⁠. Basics: powers

  2. Substitute t=−x2t = -x^2

    Let t=−x2t = -x^2⁠. Then dt=−2x dxdt = -2x\,dx⁠, so

    x dx=−12 dt.x\,dx = -\frac12\,dt.

    Therefore

    ∫x e−x2 dx=∫et(−12 dt)=−12∫et dt.\int x\,e^{-x^2}\,dx = \int e^t\left(-\frac12\,dt\right) = -\frac12\int e^t\,dt.
    We solved dt=−2x dxdt = -2x\,dx for exactly the product x dxx\,dx that the integrand contains. Choosing t=−x2t = -x^2 (rather than t=x2t = x^2⁠) makes the new integrand the plain ete^t of the standard formula, with no minus sign in the exponent.
    Change of variable, Section 7.3.1; constants come outside, Property (IV), Section 7.2.1. Lesson: integration by substitution

  3. Integrate and return to xx

    −12∫et dt=−12et+C=−12e−x2+C=−12ex2+C.\begin{aligned} -\frac12\int e^t\,dt &= -\frac12 e^t + C \\ &= -\frac12 e^{-x^2} + C \\ &= -\frac{1}{2e^{x^2}} + C. \end{aligned}
    The last line rewrites e−x2e^{-x^2} as 1ex2\frac{1}{e^{x^2}}⁠, which is how the textbook states the answer. The two forms are the same function.
    Standard formula (xi), Section 7.2: ∫ex dx=ex+C\int e^x\,dx = e^x + C⁠.

  4. Check: differentiate the answer back

    By the chain rule (outer eue^u⁠, inner u=−x2u = -x^2 with u′=−2xu' = -2x⁠):

    ddx[−12e−x2]=−12⋅e−x2⋅(−2x)=x e−x2=xex2.\begin{aligned} \frac{d}{dx}\left[-\frac12 e^{-x^2}\right] &= -\frac12\cdot e^{-x^2}\cdot(-2x) \\ &= x\,e^{-x^2} = \frac{x}{e^{x^2}}. \end{aligned}

    This is the integrand, valid for every real xx⁠.

    The inner derivative −2x-2x times the constant −12-\frac12 gives exactly xx⁠, the factor in front of the exponential. Getting the integrand back confirms the answer (Property (I), Section 7.2.1).
    Chain rule, Theorem 4, Section 5.3.1; ddxex=ex\frac{d}{dx}e^x = e^x⁠, Theorem 5, Section 5.4.

  5. The same with t=x2t = x^2

    If instead we put t=x2t = x^2⁠, then x dx=12 dtx\,dx = \frac12\,dt and the integral is 12∫e−t dt\frac12\int e^{-t}\,dt⁠. Since ddt(−e−t)=e−t\frac{d}{dt}\left(-e^{-t}\right) = e^{-t}⁠, this equals −12e−t+C=−12e−x2+C-\frac12 e^{-t} + C = -\frac12 e^{-x^2} + C⁠, the same answer.

    Either substitution works. The one in Step 2 is a little quicker because it lands directly on the standard formula ∫et dt\int e^t\,dt⁠.
Result ∫xex2 dx=−12ex2+C,x∈R.\int \frac{x}{e^{x^2}}\,dx = -\frac{1}{2e^{x^2}} + C, \qquad x \in \R.

(Textbook answer: −12ex2+C-\dfrac{1}{2e^{x^2}} + C⁠.)

Check it with numbers

Let F(x)=−12e−x2F(x) = -\frac12 e^{-x^2} and look near x=1x = 1 with h=0.001h = 0.001⁠:

  • F(1.001)≈−0.1835720252F(1.001) \approx -0.1835720252 and F(0.999)≈−0.1843077838F(0.999) \approx -0.1843077838⁠.
  • Slope estimate: F(1.001)−F(0.999)0.002\dfrac{F(1.001) - F(0.999)}{0.002} ≈0.367879\approx 0.367879⁠.
  • Integrand at x=1x = 1⁠: 1e≈0.367879\dfrac{1}{e} \approx 0.367879⁠.

They agree.

Common mistakes
  1. Losing the minus sign. The answer is negative: −12ex2-\frac{1}{2e^{x^2}}⁠. Differentiating +12ex2+\frac{1}{2e^{x^2}} gives −xex2-\frac{x}{e^{x^2}}⁠.
  2. Trying ∫e−x2 dx\int e^{-x^2}\,dx alone. Without the factor xx there is no derivative to absorb, and that integral has no elementary antiderivative (Remark (ii), Section 7.2). The xx is what makes this question work.
  3. Treating ex2e^{x^2} as (ex)2(e^x)^2⁠. (ex)2=e2x(e^x)^2 = e^{2x}⁠, a different function.
What to take away
  1. ∫g′(x) eg(x) dx=eg(x)+C\displaystyle\int g'(x)\,e^{g(x)}\,dx = e^{g(x)} + C⁠. Look for the derivative of the exponent as a factor.
  2. Rewrite quotients with negative exponents first; the pattern is then easier to see.

Every fact used, and where it comes from

FactUsed inSource
1ea=e−a\frac{1}{e^a} = e^{-a}Steps 1, 3Laws of indices
Change of variable (substitution)Steps 2, 5Section 7.3.1 (Lesson: integration by substitution)
∫kf=k∫f\int k f = k\int fStep 2Property (IV), Section 7.2.1
∫et dt=et+C\int e^t\,dt = e^t + CStep 3Standard formula (xi), Section 7.2
ddxex=ex\frac{d}{dx}e^x = e^x⁠; chain ruleSteps 4, 5Theorem 5, Section 5.4; Theorem 4, Section 5.3.1
Some functions have no elementary antiderivativeCommon mistake 2Remark (ii), Section 7.2
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