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Chapter 7 · Integrals

Exercise 7.8 · Q15: ∫01x ex2 dx\int_0^1 x\,e^{x^2}\,dx

Exercise 7.8Fundamental theorem of calculus · Section 7.8
The question

Evaluate the definite integral

∫01x ex2 dx.\int_0^1 x\,e^{x^2}\,dx.
Exam answer

Let I=∫01x ex2 dxI = \displaystyle\int_0^1 x\,e^{x^2}\,dx⁠.

Put x2=tx^2 = t⁠. Then 2x dx=dt2x\,dx = dt⁠, so x dx=12 dtx\,dx = \frac12\,dt⁠.

When x=0x = 0⁠, t=0t = 0⁠; when x=1x = 1⁠, t=1t = 1⁠.

I=12∫01et dt=12[et]01=12(e1−e0)\begin{aligned} I &= \frac12\int_0^1 e^t\,dt = \frac12\Big[e^t\Big]_0^1 \\ &= \frac12\left(e^1 - e^0\right) \end{aligned}

∴ I=12(e−1)\therefore\ I = \dfrac12(e - 1)⁠.

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

The one idea this problem is built on

The factor xx is half the derivative of the exponent x2x^2⁠. So the substitution t=x2t = x^2 turns the integrand into 12et\frac12 e^t⁠, whose antiderivative is 12et\frac12 e^t⁠. Then Theorem 2 (Section 7.8.3) gives F(1)−F(0)F(1) - F(0)⁠.

Given, to find, and the plan
Given
The integrand f(x)=x ex2f(x) = x\,e^{x^2} on [0,1][0, 1]⁠.
To find
The exact value of ∫01x ex2dx\int_0^1 x\,e^{x^2}dx⁠, and a decimal.
Plan
(1) Check continuity. (2) Substitute t=x2t = x^2 to find FF⁠. (3) Differentiate FF back. (4) Compute F(1)−F(0)F(1) - F(0)⁠.

Basics you need

1Second fundamental theorem

∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a) when ff is continuous on [a,b][a, b] and F′=fF' = f⁠. Lesson: definite integrals and the fundamental theorem

2Substitution

t=g(x)t = g(x)⁠, dt=g′(x) dxdt = g'(x)\,dx⁠; the whole integrand must be rewritten in tt⁠. Basics: substitution (the chain rule backwards)

3The exponential

∫et dt=et+C\int e^t\,dt = e^t + C⁠, and e0=1e^0 = 1⁠. Basics: logarithms and 1/x

Solution

  1. Check continuity on [0,1][0, 1]

    xx is a polynomial, and ex2e^{x^2} is the composition of the continuous functions x↦x2x \mapsto x^2 and u↦euu \mapsto e^u⁠. Their product is continuous for every real xx⁠, in particular on [0,1][0, 1]⁠.

    Theorem 2 requires continuity on the whole closed interval. Exponentials and polynomials are continuous everywhere, so there is nothing to exclude.
    Products and compositions of continuous functions are continuous (Chapter 5). Lesson: the fundamental theorem

  2. Substitute t=x2t = x^2

    Then dt=2x dxdt = 2x\,dx⁠, so x dx=12 dtx\,dx = \frac12\,dt⁠, and

    ∫x ex2dx=12∫et dt=12et+C=12ex2+C.\int x\,e^{x^2}dx = \frac12\int e^t\,dt = \frac12 e^t + C = \frac12 e^{x^2} + C.

    Take F(x)=12ex2F(x) = \frac12 e^{x^2}⁠.

    The exponent is x2x^2 and its derivative 2x2x appears (up to the factor 2) as the other factor. After the substitution only ete^t is left.
    Standard formula (xi), Section 7.2: ∫ex dx=ex+C\int e^x\,dx = e^x + C⁠. Lesson: integration by substitution

  3. Check: differentiate FF back

    F′(x)=12 ex2⋅2x=x ex2. ✓F'(x) = \frac12\,e^{x^2}\cdot 2x = x\,e^{x^2}.\ \checkmark
    Chain rule with outer function 12eu\frac12 e^u and inner function u=x2u = x^2⁠, dudx=2x\frac{du}{dx} = 2x⁠. The 2 cancels the 12\frac12⁠.
    ddueu=eu\frac{d}{du}e^u = e^u⁠. Basics: integration undoes differentiation

  4. Evaluate F(1)−F(0)F(1) - F(0)

    I=12e12−12e02=12e−12⋅1=12(e−1)≈0.8591409.\begin{aligned}I &= \frac12 e^{1^2} - \frac12 e^{0^2} = \frac12 e - \frac12\cdot 1 \\ &= \frac12(e - 1) \approx 0.8591409.\end{aligned}

    With new limits instead: x=0x = 0 gives t=0t = 0⁠, x=1x = 1 gives t=1t = 1⁠, and 12∫01et dt=12(e−1)\frac12\int_0^1 e^t\,dt = \frac12(e - 1)⁠.

    0.5 1 1 2 e x y y = x·e^(x²) area (e − 1)/2 ≈ 0.859
    The shaded region under y=x ex2y = x\,e^{x^2} from 0 to 1 has area 12(e−1)≈0.859\frac12(e - 1) \approx 0.859⁠. The curve rises from 0 to e≈2.718e \approx 2.718⁠, so the area is less than the rectangle 1×e1 \times e⁠, as it should be.
    Upper limit first: F(1)=e2F(1) = \frac{e}{2}⁠, F(0)=e02=12F(0) = \frac{e^0}{2} = \frac12⁠. Do not forget the lower limit: e0=1e^0 = 1⁠, not 0. The new limits are legitimate because t=x2t = x^2 is increasing on [0,1][0, 1]⁠, so tt runs once from 0 to 1.
    e0=1e^0 = 1⁠, e≈2.7182818e \approx 2.7182818⁠. Basics: area as a sum of thin strips
Result ∫01x ex2dx=12(e−1)≈0.8591409.\int_0^1 x\,e^{x^2}dx = \frac12(e - 1) \approx 0.8591409.

(Textbook answer: 12(e−1)\frac12(e - 1)⁠.)

Check it with numbers

A numerical integration of x ex2x\,e^{x^2} over [0,1][0, 1] gives about 0.8591409, and e−12≈1.71828182=0.8591409\frac{e - 1}{2} \approx \frac{1.7182818}{2} = 0.8591409⁠. They agree.

Common mistakes
  1. Taking F(0)=0F(0) = 0⁠. F(0)=12e0=12F(0) = \frac12 e^0 = \frac12⁠; forgetting it gives e2\frac{e}{2} instead of e−12\frac{e - 1}{2}⁠.
  2. Writing ∫ex2dx\int e^{x^2}dx as ex22x\frac{e^{x^2}}{2x}⁠. Dividing by the inner derivative only works when that derivative is a constant; here the factor xx in front is what makes the substitution work.
  3. Dropping the 12\frac12⁠. dt=2x dxdt = 2x\,dx⁠, so x dx=12dtx\,dx = \frac12 dt⁠.
What to take away
  1. ∫g′(x) eg(x)dx=eg(x)+C\int g'(x)\,e^{g(x)}dx = e^{g(x)} + C⁠: look for the derivative of the exponent as a factor.
  2. Evaluate the antiderivative at both limits, even when one of them is 0.
  3. Exact answers keep ee as a symbol; give a decimal as well.

Every fact used, and where it comes from

FactUsed inSource
∫abf=F(b)−F(a)\int_a^b f = F(b) - F(a) for continuous ffSteps 1, 4Theorem 2, Section 7.8.3 (Lesson)
Substitution t=x2t = x^2Step 2Section 7.3.1 (Lesson)
∫ex dx=ex+C\int e^x\,dx = e^x + CStep 2Standard formula (xi), Section 7.2
Chain ruleStep 3Chapter 5 (Basics)
Changing the limits after a substitutionStep 4Section 7.9
Definite integral as areaFigureSection 7.8.1 (Basics)
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