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Chapter 7 · Integrals

Exercise 7.8 · Q3: ∫12(4x3−5x2+6x+9) dx\int_1^2(4x^3 - 5x^2 + 6x + 9)\,dx

Exercise 7.8Fundamental theorem of calculus · Section 7.8
The question

Evaluate the definite integral

∫12(4x3−5x2+6x+9)dx.\int_1^2\left(4x^3 - 5x^2 + 6x + 9\right)dx.
Exam answer

Let I=∫12(4x3−5x2+6x+9)dxI = \displaystyle\int_1^2\left(4x^3 - 5x^2 + 6x + 9\right)dx⁠.

I=[x4−5x33+3x2+9x]12=(16−403+12+18)−(1−53+3+9)=983−343\begin{aligned} I &= \left[x^4 - \frac{5x^3}{3} + 3x^2 + 9x\right]_1^2 \\ &= \left(16 - \frac{40}{3} + 12 + 18\right) \\ &\quad - \left(1 - \frac53 + 3 + 9\right) \\ &= \frac{98}{3} - \frac{34}{3} \end{aligned}

∴ I=643\therefore\ I = \dfrac{64}{3}⁠.

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) for continuous ff and any antiderivative FF⁠. A polynomial is integrated term by term with the power rule, so the only real work is careful arithmetic with fractions when F(2)F(2) and F(1)F(1) are computed.

Given, to find, and the plan
Given
f(x)=4x3−5x2+6x+9f(x) = 4x^3 - 5x^2 + 6x + 9 on [1,2][1, 2]⁠.
To find
The exact value of ∫12f(x) dx\int_1^2 f(x)\,dx⁠.
Plan
(1) Continuity. (2) Antiderivative term by term, then differentiate it back. (3) F(2)F(2)⁠, F(1)F(1) and their difference, with exact fractions. (4) Check the size: ff rises from 14 to 33 on [1,2][1, 2]⁠.

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

2Power rule

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

3Sum and constant-multiple rules

∫[k1f+k2g] dx=k1∫f dx+k2∫g dx\int[k_1 f + k_2 g]\,dx = k_1\int f\,dx + k_2\int g\,dx (Properties (III) to (V), Section 7.2.1): constants come out, terms are integrated one at a time. Basics: integration undoes differentiation

Solution

  1. Check continuity on [1,2][1, 2]

    ff is a polynomial, so it is continuous at every real number, in particular on [1,2][1, 2]⁠.

    Theorem 2 needs ff continuous on the closed interval (Remark (iv) after Theorem 2). Polynomials always are.
    Every polynomial is continuous: Chapter 5, Example 14. Lesson: the fundamental theorem

  2. Integrate term by term

    Apply the power rule to each term, keeping the constant in front:

    ∫4x3 dx=4⋅x44=x4,∫5x2 dx=5⋅x33=5x33,∫6x dx=6⋅x22=3x2,∫9 dx=9x.\begin{aligned}\int 4x^3\,dx &= 4\cdot\frac{x^4}{4} = x^4, \\ \int 5x^2\,dx &= 5\cdot\frac{x^3}{3} = \frac{5x^3}{3}, \\ \int 6x\,dx &= 6\cdot\frac{x^2}{2} = 3x^2, \\ \int 9\,dx &= 9x.\end{aligned}

    So, with C=0C = 0⁠,

    F(x)=x4−5x33+3x2+9x.F(x) = x^4 - \frac{5x^3}{3} + 3x^2 + 9x.
    The power rule raises each power by 1 and divides by the new power. Constants such as 4, 5, 6 stay as multipliers (constant-multiple rule). The constant C is dropped because it cancels in F(2)−F(1)F(2) - F(1)⁠.
    Standard formula (i), Section 7.2, with n=3,2,1,0n = 3, 2, 1, 0⁠. Basics: the power rule

  3. Check: differentiate FF back

    F′(x)=4x3−5⋅3x23+6x+9=4x3−5x2+6x+9=f(x).\begin{aligned}F'(x) &= 4x^3 - \frac{5\cdot 3x^2}{3} + 6x + 9 \\ &= 4x^3 - 5x^2 + 6x + 9 = f(x).\end{aligned}
    Every term returns exactly, so FF is a correct antiderivative for all real xx⁠.
    ddxxn=nxn−1\frac{d}{dx}x^n = nx^{n-1}⁠. Basics: integration undoes differentiation

  4. Compute F(2)F(2)

    F(2)=24−5⋅233+3⋅22+9⋅2=16−403+12+18=46−403=138−403=983.\begin{aligned}F(2) &= 2^4 - \frac{5\cdot 2^3}{3} + 3\cdot 2^2 + 9\cdot 2 \\ &= 16 - \frac{40}{3} + 12 + 18 \\ &= 46 - \frac{40}{3} = \frac{138 - 40}{3} = \frac{98}{3}.\end{aligned}
    Group the whole numbers first (16+12+18=4616 + 12 + 18 = 46⁠), then write 46 as 1383\frac{138}{3} to subtract the fraction.
    24=162^4 = 16⁠, 23=82^3 = 8⁠, 22=42^2 = 4⁠. Basics: powers

  5. Compute F(1)F(1) and subtract

    F(1)=1−53+3+9=13−53=343.F(1) = 1 - \frac53 + 3 + 9 = 13 - \frac53 = \frac{34}{3}.

    Therefore

    ∫12f(x) dx=F(2)−F(1)=983−343=643≈21.3333333.\begin{aligned}\int_1^2 f(x)\,dx &= F(2) - F(1) = \frac{98}{3} - \frac{34}{3} \\ &= \frac{64}{3} \approx 21.3333333.\end{aligned}
    Upper limit minus lower limit. Both values have denominator 3, so the subtraction is immediate.
    [F(x)]12=F(2)−F(1)\big[F(x)\big]_1^2 = F(2) - F(1)⁠. Lesson: the fundamental theorem

  6. Does the size make sense?

    f(1)=4−5+6+9=14f(1) = 4 - 5 + 6 + 9 = 14 and f(2)=32−20+12+9=33f(2) = 32 - 20 + 12 + 9 = 33⁠. The derivative f′(x)=12x2−10x+6f'(x) = 12x^2 - 10x + 6 has discriminant 100−288=−188<0100 - 288 = -188 < 0 and positive leading coefficient, so f′(x)>0f'(x) > 0 for every xx⁠: ff increases. On [1,2][1, 2]⁠, then, 14≤f(x)≤3314 \le f(x) \le 33⁠, and on a base of width 1:

    14<643≈21.33<33.14 < \frac{64}{3} \approx 21.33 < 33.
    1 2 10 20 30 40 x y y = 4x³ − 5x² + 6x + 9 (1, 14) (2, 33) area 64/3 ≈ 21.33
    The shaded region under y=4x3−5x2+6x+9y = 4x^3 - 5x^2 + 6x + 9 from 1 to 2 has area 643≈21.33\frac{64}{3} \approx 21.33⁠. The curve rises from 14 to 33 on this interval, so the area lies between those two rectangle areas.
    Since f>0f > 0 on [1,2][1, 2]⁠, the integral is the shaded area (Section 7.8.1), and it must lie between the areas of the lowest and highest rectangles on the same base.
    A quadratic ax2+bx+cax^2 + bx + c with a>0a > 0 and b2−4ac<0b^2 - 4ac < 0 is positive for every xx⁠. Basics: area as a sum of thin strips
Result ∫12(4x3−5x2+6x+9)dx=643≈21.3333333.\int_1^2\left(4x^3 - 5x^2 + 6x + 9\right)dx = \frac{64}{3} \approx 21.3333333.

(Textbook answer: 643\frac{64}{3}⁠, the same.)

Check it with numbers

Numerical integration of the polynomial from 1 to 2 (thin strips, no antiderivative used) gives 21.333333321.3333333⁠, and 643=21.3333333…\frac{64}{3} = 21.3333333\ldots⁠: they agree.

Common mistakes
  1. Forgetting to divide by the new power. ∫5x2 dx=5x33\int 5x^2\,dx = \frac{5x^3}{3}⁠, not 5x35x^3⁠.
  2. Dropping the 9x9x⁠. The constant term 9 integrates to 9x9x⁠, which contributes 18−9=918 - 9 = 9 to the answer.
  3. Fraction slips. Write F(2)F(2) and F(1)F(1) over the common denominator 3 before subtracting.
What to take away
  1. Polynomials are integrated term by term with the power rule; they are continuous everywhere, so any limits are allowed.
  2. Evaluate F(b)F(b) and F(a)F(a) separately, as exact fractions, then subtract.
  3. The smallest and largest values of ff on [a,b][a, b]⁠, times b−ab - a⁠, bound the integral.

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, 5Theorem 2, Section 7.8.3 (Lesson)
Polynomials are continuousStep 1Chapter 5, Section 5.2 (Example 14)
∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C⁠, n≠−1n \ne -1Step 2Standard formula (i), Section 7.2 (Basics)
Sum and constant-multiple rulesStep 2Properties (III) to (V), Section 7.2.1
Differentiation undoes integrationStep 3Property (I), Section 7.2.1 (Basics)
Definite integral as areaStep 6Section 7.8.1 (Basics)
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