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Chapter 7 · Integrals

Exercise 7.6 · Q23 (MCQ): ∫x2ex3 dx\int x^2e^{x^3}\,dx

Exercise 7.6Integration by substitution · Section 7.3.1
The question

Choose the correct answer.

∫x2ex3 dx equals\int x^2 e^{x^3}\,dx \text{ equals}

(A) 13ex3+C\frac13 e^{x^3} + C   (B) 13ex2+C\frac13 e^{x^2} + C   (C) 12ex3+C\frac12 e^{x^3} + C   (D) 12ex2+C\frac12 e^{x^2} + C

Exam answer

Put x3=tx^3 = t⁠. Then 3x2 dx=dt3x^2\,dx = dt⁠.

∫x2ex3 dx=13∫et dt=13et+C\begin{aligned} \int x^2e^{x^3}\,dx &= \frac13\int e^t\,dt \\ &= \frac13e^t + C \end{aligned}

Answer: (A) 13ex3+C\dfrac13 e^{x^3} + C

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

The one idea this problem is built on

The factor x2x^2 is, up to the constant 3, the derivative of the exponent x3x^3⁠. So the substitution t=x3t = x^3 turns the integral into 13∫et dt\frac13\int e^t\,dt⁠. Although the question sits in the integration by parts exercise, substitution is the natural method here.

Given, to find, and the plan
Given
x2ex3x^2 e^{x^3}⁠, defined and continuous for every real xx⁠, and four candidate answers.
To find
∫x2ex3 dx\int x^2 e^{x^3}\,dx and the matching option.
Plan
(1) Substitute t=x3t = x^3⁠. (2) Integrate and substitute back. (3) Differentiate back. (4) Examine every option.

Basics you need

1Integration by substitution

If t=g(x)t = g(x) then dt=g′(x) dxdt = g'(x)\,dx⁠; an integrand of the form f(g(x)) g′(x)f(g(x))\,g'(x) becomes f(t)f(t) (Section 7.3.1). Basics: substitution (the chain rule backwards)

2∫et dt=et+C\int e^t\,dt = e^t + C

Standard formula (xi), Section 7.2. Lesson: integration by substitution

3Chain rule

ddxeg(x)=eg(x) g′(x)\frac{d}{dx}e^{g(x)} = e^{g(x)}\,g'(x)⁠. Basics: integration undoes differentiation

Solution

  1. Substitute t=x3t = x^3

    Put t=x3t = x^3⁠. Then dt=3x2 dxdt = 3x^2\,dx⁠, so x2 dx=13 dtx^2\,dx = \frac13\,dt⁠, and

    ∫x2ex3 dx=∫et⋅13 dt=13∫et dt.\int x^2 e^{x^3}\,dx = \int e^t\cdot\frac13\,dt = \frac13\int e^t\,dt.
    The exponent x3x^3 is the "inside" function, and its derivative 3x23x^2 is present in the integrand apart from the constant 3. A missing constant is never a problem: it is moved outside the integral. The substitution is valid for every real xx⁠.
    Substitution (Section 7.3.1). Basics: substitution (the chain rule backwards)

  2. Integrate and substitute back

    13∫et dt=13et+C=13ex3+C.\frac13\int e^t\,dt = \frac13 e^t + C = \frac13 e^{x^3} + C.
    ete^t is its own antiderivative. Replacing tt by x3x^3 returns the answer in terms of xx⁠.
    ∫et dt=et+C\int e^t\,dt = e^t + C (formula (xi)). Lesson: integration by substitution

  3. Check: differentiate the answer

    ddx[13ex3]=13ex3⋅3x2=x2ex3,\frac{d}{dx}\left[\frac13 e^{x^3}\right] = \frac13 e^{x^3}\cdot 3x^2 = x^2 e^{x^3},

    the integrand, for every real xx⁠.

    Chain rule: outer function eue^u⁠, inner function u=x3u = x^3 with derivative 3x23x^2⁠. The 3 cancels the 13\frac13⁠.
    Chain rule (Chapter 5, Section 5.3.1). Basics: integration undoes differentiation

  4. Examine every option

    An option is correct exactly when its derivative is x2ex3x^2 e^{x^3}⁠. Differentiate each (chain rule):

    (A) ddx13ex3=x2ex3\frac{d}{dx}\frac13 e^{x^3} = x^2 e^{x^3}⁠. Correct (Step 3).

    (B) ddx13ex2=23x ex2\frac{d}{dx}\frac13 e^{x^2} = \frac23 x\,e^{x^2}⁠. Wrong exponent; at x=1x = 1 it gives 23e\frac23 e⁠, not ee⁠.

    (C) ddx12ex3=32x2ex3\frac{d}{dx}\frac12 e^{x^3} = \frac32 x^2 e^{x^3}⁠. Right shape, but 32\frac32 times too big.

    (D) ddx12ex2=x ex2\frac{d}{dx}\frac12 e^{x^2} = x\,e^{x^2}⁠. Wrong exponent and wrong power of xx⁠; at x=2x = 2 it gives 2e42e^4⁠, not 4e84e^8⁠.

    Differentiating the options is the safest way to answer an integration MCQ: it needs only the chain rule, and an option is right only if it gives back the integrand exactly.
    Chain rule. Basics: integration undoes differentiation
Result ∫x2ex3 dx=13ex3+C,\int x^2 e^{x^3}\,dx = \frac13 e^{x^3} + C,

valid for every real xx⁠. The correct option is (A). (Textbook answer: A.)

Check it with numbers

On [0,1][0, 1]⁠:

F(1)−F(0)=13e−13=e−13=0.5727606095…\begin{aligned} F(1) - F(0) &= \frac13 e - \frac13 \\ &= \frac{e - 1}{3} = 0.5727606095\ldots \end{aligned}

Numerical integration of x2ex3x^2 e^{x^3} over [0,1][0, 1] gives 0.5727606095…0.5727606095\ldots⁠, the same.

Common mistakes
  1. Forgetting the 13\frac13⁠: x2 dxx^2\,dx is 13 dt\frac13\,dt⁠, not dtdt⁠.
  2. Trying integration by parts because of the exercise heading: with x2x^2 first and ex3e^{x^3} second you would need ∫ex3 dx\int e^{x^3}\,dx⁠, which has no elementary formula (a known result, not proved in the book; compare Remark (ii) after Example 4).
  3. Changing the exponent to x2x^2 (options (B), (D)): the exponent stays x3x^3⁠; only the factor in front changes.
What to take away
  1. ∫g′(x) eg(x) dx=eg(x)+C\int g'(x)\,e^{g(x)}\,dx = e^{g(x)} + C⁠; fix constant factors by multiplying and dividing.
  2. Check MCQ options by differentiating them.

Every fact used, and where it comes from

FactUsed inSource
Integration by substitutionStep 1Section 7.3.1 (Lesson: integration by substitution)
∫et dt=et+C\int e^t\,dt = e^t + CStep 2Standard formula (xi), Section 7.2
Chain ruleSteps 3, 4Chapter 5, Section 5.3.1
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