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Chapter 7 · Integrals

Exercise 7.1 · Q9: ∫(2x2+ex) dx\int (2x^2 + e^x)\,dx

Exercise 7.1Properties of indefinite integrals · Section 7.2.1
The question

Find the following integral:

∫(2x2+ex)dx\int\left(2x^2 + e^x\right)dx
Exam answer

Let I=∫(2x2+ex)dxI = \displaystyle\int\left(2x^2 + e^x\right)dx⁠.

I=2∫x2 dx+∫ex dxI = 2\int x^2\,dx + \int e^x\,dx

∴ I=2x33+ex+C\therefore\ I = \dfrac{2x^3}{3} + e^x + C⁠.

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

The one idea this problem is built on

Two different kinds of term, each with its own standard formula: a power of xx (power rule) and the exponential exe^x⁠, which is its own anti derivative. Split the integral and use each formula on its own term.

Given, to find, and the plan
Given
The integrand 2x2+ex2x^2 + e^x⁠, defined for every real xx⁠.
To find
The indefinite integral.
Plan
(1) Split and take out the constant 22⁠. (2) Power rule for x2x^2⁠. (3) Standard formula for exe^x⁠. (4) Assemble with one CC⁠. (5) Differentiate back to check.

Basics you need

1Power rule for integrals

∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C⁠, n≠−1n \ne -1 (item (i), Section 7.2). Basics: powers of x and the power rule

2The exponential

ddxex=ex\frac{d}{dx}e^x = e^x⁠, so ∫ex dx=ex+C\int e^x\,dx = e^x + C (item (xi), Section 7.2). Basics: integration undoes differentiation

3Term by term

Properties (III) and (IV), Section 7.2.1: split sums and pull out constants. Lesson: integration as anti differentiation

Solution

  1. Split and take out the constant

    ∫(2x2+ex)dx=2∫x2 dx+∫ex dx.\int\left(2x^2 + e^x\right)dx = 2\int x^2\,dx + \int e^x\,dx.
    Property (III) splits the sum; Property (IV) moves the constant 22 outside the first integral. Now each integral matches one standard formula.
    Properties (III) and (IV), Section 7.2.1. Lesson: integration as anti differentiation

  2. Power rule for x2x^2

    2∫x2 dx=2⋅x33=2x33(+ const).2\int x^2\,dx = 2\cdot\frac{x^{3}}{3} = \frac{2x^3}{3}\quad(+\text{ const}).
    With n=2n = 2⁠, the power rule raises the exponent to 33 and divides by 33⁠. The factor 22 then multiplies the result.
    Standard formulae, item (i), Section 7.2. Basics: the power rule

  3. The exponential term

    ∫ex dx=ex(+ const).\int e^x\,dx = e^x\quad(+\text{ const}).
    Since ddxex=ex\frac{d}{dx}e^x = e^x⁠, the function exe^x is an anti derivative of itself. Note the contrast with the power rule: exe^x is not a power of xx⁠, so we do not write ex+1x+1\frac{e^{x+1}}{x+1}⁠.
    Standard formulae, item (xi), Section 7.2; Theorem 5, Section 5.4 (Chapter 5). Basics: integration undoes differentiation

  4. Assemble the answer

    ∫(2x2+ex)dx=2x33+ex+C.\int\left(2x^2 + e^x\right)dx = \frac{2x^3}{3} + e^x + C.
    The constants from the two pieces combine into one arbitrary CC⁠.
    Note after Example 2. Example 2 (iii) has the same mix of a power and 2ex2e^x⁠. Example 2

  5. Check: differentiate the answer back

    ddx(2x33+ex+C)=2⋅3x23+ex=2x2+ex.  ✓\begin{aligned} \frac{d}{dx}\left(\frac{2x^3}{3} + e^x + C\right) &= \frac{2\cdot 3x^2}{3} + e^x \\ &= 2x^2 + e^x. \;\checkmark \end{aligned}
    The derivative is the integrand for every real xx⁠.
    Property (I), Section 7.2.1. Basics: integration undoes differentiation
Result ∫(2x2+ex)dx=2x33+ex+C,\int\left(2x^2 + e^x\right)dx = \frac{2x^3}{3} + e^x + C,

valid for every real xx⁠. (Textbook answer: 23x3+ex+C\frac23 x^3 + e^x + C⁠, the same expression.)

Check it with numbers

At x=1x = 1 with h=0.001h = 0.001⁠, for F(x)=2x33+exF(x) = \frac{2x^3}{3} + e^x⁠:

F(1.001)−F(0.999)0.002=3.389670137−3.3802335710.002≈4.718283.\begin{aligned} &\frac{F(1.001) - F(0.999)}{0.002} \\ &= \frac{3.389670137 - 3.380233571}{0.002} \\ &\approx 4.718283. \end{aligned}

The integrand at x=1x = 1 is 2+e≈4.7182822 + e \approx 4.718282⁠. They agree to six significant figures.

Common mistakes
  1. Power rule on exe^x⁠. ex+1x+1\frac{e^{x+1}}{x+1} is wrong: the power rule needs a fixed exponent on the variable xx⁠, and here the variable is in the exponent.
  2. Losing the 2. ∫2x2 dx=2x33\int 2x^2\,dx = \frac{2x^3}{3}⁠, not x33\frac{x^3}{3}⁠.
  3. Dividing by the old power. 2x32\frac{2x^3}{2} divides by the old exponent; divide by the new exponent 33⁠.
What to take away
  1. Recognise each term's type (power, exponential, trigonometric) and use its own formula.
  2. ∫ex dx=ex+C\int e^x\,dx = e^x + C⁠: the exponential is unchanged by integration.

Every fact used, and where it comes from

FactUsed inSource
Split sums, take out constantsStep 1Properties (III), (IV), Section 7.2.1
∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C⁠, n≠−1n \ne -1Step 2Standard formulae, item (i), Section 7.2
∫ex dx=ex+C\int e^x\,dx = e^x + CStep 3Standard formulae, item (xi), Section 7.2
One constant of integrationStep 4Note after Example 2, Section 7.2.1
Differentiation and integration undo each otherStep 5Property (I), Section 7.2.1
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