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Chapter 7 · Integrals

Exercise 7.8 · Q6: ∫45ex dx\int_4^5 e^x\,dx

Exercise 7.8Fundamental theorem of calculus · Section 7.8
The question

Evaluate the definite integral

∫45ex dx.\int_4^5 e^x\,dx.
Exam answer

Let I=∫45ex dxI = \displaystyle\int_4^5 e^x\,dx⁠.

I=[ex]45=e5−e4I = \Big[e^x\Big]_4^5 = e^5 - e^4

∴ I=e4(e−1)\therefore\ I = e^4(e - 1)⁠.

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

The one idea this problem is built on

The second fundamental theorem (Theorem 2, Section 7.8.3): ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)⁠. The function exe^x is its own derivative, so it is its own antiderivative, and the integral is simply e5−e4e^5 - e^4⁠. Taking out the common factor e4e^4 gives the book's form e4(e−1)e^4(e - 1)⁠.

Given, to find, and the plan
Given
f(x)=exf(x) = e^x on [4,5][4, 5]⁠.
To find
The exact value of ∫45ex dx\int_4^5 e^x\,dx⁠.
Plan
(1) Continuity. (2) Antiderivative exe^x⁠, checked by differentiating. (3) F(5)−F(4)F(5) - F(4)⁠, factorised. (4) Size check: exe^x rises from e4e^4 to e5e^5 on [4,5][4, 5]⁠.

Basics you need

1Second fundamental theorem

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

2Integral of exe^x

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

3Index law

e5=e4+1=e4⋅ee^5 = e^{4+1} = e^4\cdot e⁠, so e5−e4=e4(e−1)e^5 - e^4 = e^4(e - 1)⁠. Basics: powers and index laws

Solution

  1. Check continuity on [4,5][4, 5]

    The exponential function exe^x is defined and continuous for every real xx⁠, so it is continuous on [4,5][4, 5]⁠.

    Theorem 2 needs a continuous integrand on the closed interval (Remark (iv) after Theorem 2). exe^x has no gaps or jumps anywhere.
    The exponential function is continuous (Chapter 5, Section 5.4). Lesson: the fundamental theorem

  2. Find an antiderivative and check it

    By standard formula (xi), ∫ex dx=ex+C\int e^x\,dx = e^x + C⁠. Take

    F(x)=ex.F(x) = e^x.

    Check: F′(x)=ddxex=ex=f(x)F'(x) = \frac{d}{dx}e^x = e^x = f(x) for every real xx⁠.

    exe^x is the one function (up to a constant factor) that is its own derivative, so the antiderivative needs no work at all.
    Standard formula (xi), Section 7.2. Basics: integration undoes differentiation

  3. Evaluate F(5)−F(4)F(5) - F(4)

    ∫45ex dx=[ex]45=e5−e4=e4⋅e−e4=e4(e−1).\begin{aligned}\int_4^5 e^x\,dx &= \big[e^x\big]_4^5 = e^5 - e^4 \\ &= e^4\cdot e - e^4 = e^4(e - 1).\end{aligned}

    In decimals, e5≈148.4131591e^5 \approx 148.4131591 and e4≈54.5981500e^4 \approx 54.5981500⁠, so

    e4(e−1)=e5−e4≈93.8150091.e^4(e - 1) = e^5 - e^4 \approx 93.8150091.

    (Multiplying the rounded values 54.598150054.5981500 and e−1≈1.7182818e - 1 \approx 1.7182818 gives only 93.815007593.8150075⁠: the rounding error in e−1e - 1 is multiplied by about 55. The subtraction keeps all seven decimals right.)

    Upper minus lower. Writing e5e^5 as e4⋅ee^4\cdot e shows the common factor e4e^4⁠; taking it out gives the compact form e4(e−1)e^4(e - 1)⁠, which is how the book states the answer.
    am+n=am⋅ana^{m+n} = a^m\cdot a^n⁠. Basics: index laws

  4. Does the size make sense?

    exe^x is positive and increasing, so on [4,5][4, 5] it lies between e4e^4 and e5e^5⁠. On a base of width 1 the area is therefore between them:

    e4≈54.60<93.82<e5≈148.41.e^4 \approx 54.60 < 93.82 < e^5 \approx 148.41.
    1 2 3 4 5 50 100 150 x y (4, e⁴) (5, e⁵) y = eˣ area e⁴(e − 1) ≈ 93.82
    The shaded region under y=exy = e^x from 4 to 5 has area e4(e−1)≈93.82e^4(e - 1) \approx 93.82⁠, between the heights e4≈54.6e^4 \approx 54.6 and e5≈148.4e^5 \approx 148.4 at its two ends.
    Since ex>0e^x > 0⁠, the integral is the shaded area (Section 7.8.1). The steep climb of exe^x is why one unit of width already holds an area of almost 94.
    Definite integral as area, Section 7.8.1. Basics: area as a sum of thin strips
Result ∫45ex dx=e5−e4=e4(e−1)≈93.8150091.\int_4^5 e^x\,dx = e^5 - e^4 = e^4(e - 1) \approx 93.8150091.

(Textbook answer: e4(e−1)e^4(e - 1)⁠, the same.)

Check it with numbers

Numerical integration of exe^x from 4 to 5 (thin strips, no antiderivative used) gives 93.815009193.8150091⁠. By the formula: e5≈148.4131591e^5 \approx 148.4131591 minus e4≈54.5981500e^4 \approx 54.5981500 is 93.815009193.8150091⁠. They agree.

Common mistakes
  1. Treating exe^x like a power of x. ∫ex dx\int e^x\,dx is not ex+1x+1\frac{e^{x+1}}{x+1}⁠; the power rule is for xnx^n⁠, with the variable in the base.
  2. Writing e5−e4=e1e^5 - e^4 = e^1⁠. Exponents subtract only when you divide: e5e4=e\frac{e^5}{e^4} = e⁠, but e5−e4=e4(e−1)e^5 - e^4 = e^4(e - 1)⁠.
What to take away
  1. ∫abex dx=eb−ea\int_a^b e^x\,dx = e^b - e^a⁠.
  2. Factor out the common power: eb−ea=ea(eb−a−1)e^b - e^a = e^a\left(e^{b-a} - 1\right)⁠.
  3. For an increasing positive integrand, f(a)(b−a)<∫abf<f(b)(b−a)f(a)(b - a) < \int_a^b f < f(b)(b - a)⁠.

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, 3Theorem 2, Section 7.8.3 (Lesson)
exe^x is continuous and ddxex=ex\frac{d}{dx}e^x = e^xSteps 1, 2Chapter 5, Section 5.4
∫ex dx=ex+C\int e^x\,dx = e^x + CStep 2Standard formula (xi), Section 7.2 (Basics)
am+n=amana^{m+n} = a^m a^nStep 3Index laws (Basics)
Definite integral as areaStep 4Section 7.8.1 (Basics)
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