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Chapter 7 · Integrals

Exercise 7.2 · Q16: ∫e2x+3 dx\int e^{2x+3}\,dx

Exercise 7.2Integration by substitution · Section 7.3.1
The question

Integrate the function

e2x+3.e^{2x+3}.
Exam answer

Let I=∫e2x+3 dxI = \displaystyle\int e^{2x + 3}\,dx⁠.

Put 2x+3=t2x + 3 = t⁠. Then 2 dx=dt2\,dx = dt⁠.

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

∴ I=12e2x+3+C\therefore\ I = \dfrac12 e^{2x + 3} + C⁠.

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

The one idea this problem is built on

The exponent is the linear expression 2x+32x + 3⁠. Put t=2x+3t = 2x + 3⁠: then dt=2 dxdt = 2\,dx⁠, so dx=12 dtdx = \frac12\,dt⁠, and the integral becomes 12∫et dt\frac12\int e^t\,dt⁠. A linear inside always costs just one constant factor, 1slope\frac{1}{\text{slope}}⁠.

Given, to find, and the plan
Given
f(x)=e2x+3f(x) = e^{2x+3}⁠, defined for every real xx⁠.
To find
∫e2x+3 dx\displaystyle\int e^{2x+3}\,dx⁠.
Plan
(1) Substitute t=2x+3t = 2x + 3 and write dxdx in terms of dtdt⁠. (2) Use ∫et dt=et+C\int e^t\,dt = e^t + C⁠. (3) Return to xx⁠. (4) Differentiate back.

Basics you need

1Substitution with a linear inside

If t=ax+bt = ax + b with a≠0a \ne 0⁠, then dt=a dxdt = a\,dx⁠, so dx=1a dtdx = \frac1a\,dt⁠. Basics: substitution

2The exponential is its own derivative

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

Solution

  1. Substitute t=2x+3t = 2x + 3

    Let t=2x+3t = 2x + 3⁠. Then dtdx=2\frac{dt}{dx} = 2⁠, so

    dt=2 dx,dx=12 dt.dt = 2\,dx, \qquad dx = \frac12\,dt.

    Therefore

    ∫e2x+3 dx=∫et⋅12 dt=12∫et dt.\int e^{2x+3}\,dx = \int e^t\cdot\frac12\,dt = \frac12\int e^t\,dt.
    The standard formula only covers ete^t with the plain variable in the exponent. The substitution makes the exponent a single letter. The price is the factor 12\frac12⁠, which records that tt changes twice as fast as xx⁠.
    Change of variable, Section 7.3.1; constants come outside, Property (IV), Section 7.2.1. Lesson: integration by substitution

  2. Integrate and return to xx

    12∫et dt=12et+C=12e2x+3+C.\begin{aligned} \frac12\int e^t\,dt &= \frac12 e^t + C \\ &= \frac12 e^{2x+3} + C. \end{aligned}
    Standard formula (xi) says ete^t is its own antiderivative, and the constant 12\frac12 stays in front. Replacing tt by 2x+32x+3 puts the answer back in terms of xx⁠.
    Standard formula (xi), Section 7.2: ∫ex dx=ex+C\int e^x\,dx = e^x + C⁠. Basics: integration undoes differentiation

  3. Check: differentiate the answer back

    By the chain rule (outer eue^u⁠, inner u=2x+3u = 2x + 3 with u′=2u' = 2⁠):

    ddx[12e2x+3]=12⋅e2x+3⋅2=e2x+3.\frac{d}{dx}\left[\frac12 e^{2x+3}\right] = \frac12\cdot e^{2x+3}\cdot 2 = e^{2x+3}.

    This is the integrand, so the answer is correct. It holds for every real xx⁠, since e2x+3e^{2x+3} is defined and differentiable everywhere.

    Differentiating multiplies by the slope 22⁠; integrating must divide by it. The check shows the two cancel.
    Chain rule, Theorem 4, Section 5.3.1; ddxex=ex\frac{d}{dx}e^x = e^x⁠, Theorem 5, Section 5.4.

  4. Another way: split off the constant

    Since e2x+3=e3⋅e2xe^{2x+3} = e^3\cdot e^{2x} and e3e^3 is a constant,

    ∫e2x+3 dx=e3∫e2x dx=e3⋅12e2x+C=12e2x+3+C.\begin{aligned} \int e^{2x+3}\,dx &= e^3\int e^{2x}\,dx \\ &= e^3\cdot\frac12 e^{2x} + C \\ &= \frac12 e^{2x+3} + C. \end{aligned}
    The same answer comes out, which is a useful cross-check. It also shows that the +3+3 in the exponent only contributes the constant factor e3e^3⁠; it does not change the factor 12\frac12⁠.
    ea+b=eaebe^{a+b} = e^a e^b (laws of indices). An antiderivative of e2xe^{2x} is 12e2x\frac12 e^{2x}⁠: Exercise 7.1, Question 3.
Result ∫e2x+3 dx=12e2x+3+C,x∈R.\int e^{2x+3}\,dx = \frac12 e^{2x+3} + C, \qquad x \in \R.

(Textbook answer: 12e2x+3+C\frac12 e^{2x+3} + C⁠.)

Check it with numbers

Let F(x)=12e2x+3F(x) = \frac12 e^{2x+3} and look near x=−1x = -1 with h=0.001h = 0.001⁠:

  • F(−0.999)≈1.3618619162F(-0.999) \approx 1.3618619162 and F(−1.001)≈1.3564253489F(-1.001) \approx 1.3564253489⁠.
  • Slope estimate: F(−0.999)−F(−1.001)0.002\dfrac{F(-0.999) - F(-1.001)}{0.002} ≈2.718284\approx 2.718284⁠.
  • Integrand at x=−1x = -1⁠: e1≈2.718282e^{1} \approx 2.718282⁠.

They agree to six significant figures.

Common mistakes
  1. Multiplying by 2 instead of dividing. ∫e2x+3 dx\int e^{2x+3}\,dx is not 2e2x+3+C2e^{2x+3} + C⁠; differentiating that gives 4e2x+34e^{2x+3}⁠.
  2. Using the "power rule" on e2x+3e^{2x+3}⁠. The exponent contains the variable, so e2x+3e^{2x+3} is not a power of xx⁠; answers like e2x+42x+4\frac{e^{2x+4}}{2x+4} are wrong.
  3. Dividing by 3 or by 2x+32x+3⁠. Only the slope 22 of the linear exponent enters.
What to take away
  1. ∫eax+b dx=1aeax+b+C\displaystyle\int e^{ax+b}\,dx = \frac1a e^{ax+b} + C for a≠0a \ne 0⁠.
  2. More generally, if ∫f(t) dt=F(t)+C\int f(t)\,dt = F(t) + C⁠, then ∫f(ax+b) dx=1aF(ax+b)+C\int f(ax+b)\,dx = \frac1a F(ax+b) + C⁠. Questions 21 and 22 use this with sec⁡2\sec^2⁠.

Every fact used, and where it comes from

FactUsed inSource
Change of variable (substitution)Step 1Section 7.3.1 (Lesson: integration by substitution)
∫kf=k∫f\int k f = k\int fSteps 1, 4Property (IV), Section 7.2.1
∫et dt=et+C\int e^t\,dt = e^t + CStep 2Standard formula (xi), Section 7.2
ddxex=ex\frac{d}{dx}e^x = e^x⁠; chain ruleStep 3Theorem 5, Section 5.4; Theorem 4, Section 5.3.1
ea+b=eaebe^{a+b} = e^a e^bStep 4Laws of indices
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