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Chapter 7 · Integrals

Exercise 7.1 · Q15: ∫x (3x2+2x+3) dx\int \sqrt{x}\,(3x^2+2x+3)\,dx

Exercise 7.1Integration as anti-differentiation · Section 7.2
The question

Find the following integral:

∫x (3x2+2x+3) dx\int \sqrt{x}\,(3x^2 + 2x + 3)\,dx
Exam answer

Let I=∫x(3x2+2x+3)dxI = \displaystyle\int\sqrt{x}\left(3x^2 + 2x + 3\right)dx⁠.

I=∫(3x5/2+2x3/2+3x1/2)dx=3⋅x7/27/2+2⋅x5/25/2+3⋅x3/23/2+C\begin{aligned} I &= \int\bigg(3x^{5/2} + 2x^{3/2} \\ &\qquad + 3x^{1/2}\bigg)dx \\ &= 3\cdot\frac{x^{7/2}}{7/2} + 2\cdot\frac{x^{5/2}}{5/2} \\ &\quad + 3\cdot\frac{x^{3/2}}{3/2} + C \end{aligned}

∴ I=67x7/2+45x5/2\therefore\ I = \dfrac67x^{7/2} + \dfrac45x^{5/2} +2x3/2+C+ 2x^{3/2} + C⁠.

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

The one idea this problem is built on

As in Q14, a product cannot be integrated factor by factor. Multiply x=x1/2\sqrt{x} = x^{1/2} into the bracket: each term becomes a constant times a power of xx⁠, 3x5/2+2x3/2+3x1/23x^{5/2} + 2x^{3/2} + 3x^{1/2}⁠. Then the power formula, applied three times, does all the work.

Given, to find, and the plan
Given
The integrand f(x)=x (3x2+2x+3)f(x) = \sqrt{x}\,(3x^2 + 2x + 3)⁠.
To find
∫f(x) dx\int f(x)\,dx⁠.
Plan
(1) Note where ff is defined. (2) Expand into powers of xx⁠. (3) Integrate term by term, constants in front. (4) Differentiate the answer to check it.

Basics you need

1Integration undoes differentiation

∫f(x) dx=F(x)+C\int f(x)\,dx = F(x) + C means F′(x)=f(x)F'(x) = f(x)⁠. Basics: integration undoes differentiation

2Multiplying powers

xm⋅xn=xm+nx^m\cdot x^n = x^{m+n}⁠: x1/2⋅x2=x5/2x^{1/2}\cdot x^2 = x^{5/2} and x1/2⋅x=x3/2x^{1/2}\cdot x = x^{3/2}⁠. Basics: powers of x

3The power formula

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

4Sums and constant multiples

∫(k1f1+k2f2) dx\int (k_1 f_1 + k_2 f_2)\,dx =k1 ⁣∫f1 dx+k2 ⁣∫f2 dx= k_1\!\int f_1\,dx + k_2\!\int f_2\,dx⁠, and the same for three or more terms (Property V, Section 7.2.1). Lesson: integration as anti-differentiation

Solution

  1. Where is the integrand defined?

    Because of x\sqrt{x}⁠, the integrand is defined for x≥0x \ge 0⁠.

    The answer is valid on [0,∞)[0, \infty)⁠; at the end point x=0x = 0 the derivative check holds as a right-hand derivative. Knowing x≥0x \ge 0 also justifies the rules of exponents used next.
    The Note in Section 7.2: always keep the interval of definition in mind.

  2. Expand into powers of xx

    x (3x2+2x+3)=3x2x1/2+2x x1/2+3x1/2=3x5/2+2x3/2+3x1/2.\begin{aligned} \sqrt{x}\,(3x^2 + 2x + 3) &= 3x^{2}x^{1/2} + 2x\,x^{1/2} + 3x^{1/2} \\ &= 3x^{5/2} + 2x^{3/2} + 3x^{1/2}. \end{aligned}
    The distributive law multiplies x1/2x^{1/2} into each of the three terms. Equal bases multiply by adding exponents: 2+12=522 + \frac12 = \frac52 and 1+12=321 + \frac12 = \frac32⁠. The result is a sum of constant multiples of single powers, exactly the shape the power formula needs.
    xm⋅xn=xm+nx^m\cdot x^n = x^{m+n} for x>0x > 0⁠. Basics: powers of x

  3. Integrate term by term

    3∫x5/2 dx=3⋅x7/27/2=67x7/2,2∫x3/2 dx=2⋅x5/25/2=45x5/2,3∫x1/2 dx=3⋅x3/23/2=2x3/2.\begin{aligned} 3\int x^{5/2}\,dx &= 3\cdot\frac{x^{7/2}}{7/2} = \frac67 x^{7/2}, \\ 2\int x^{3/2}\,dx &= 2\cdot\frac{x^{5/2}}{5/2} = \frac45 x^{5/2}, \\ 3\int x^{1/2}\,dx &= 3\cdot\frac{x^{3/2}}{3/2} = 2x^{3/2}. \end{aligned}

    Adding, with a single constant,

    ∫x (3x2+2x+3) dx=67x7/2+45x5/2+2x3/2+C.\begin{aligned} &\int \sqrt{x}\,(3x^2+2x+3)\,dx \\ &\quad = \frac67 x^{7/2} + \frac45 x^{5/2} + 2x^{3/2} + C. \end{aligned}
    Property V lets the constants 3, 2, 3 come out in front and the three terms be integrated separately. For each, raise the power by 1 and divide by the new power: 3⋅27=673\cdot\frac27 = \frac67⁠, 2⋅25=452\cdot\frac25 = \frac45⁠, 3⋅23=23\cdot\frac23 = 2⁠. None of the exponents is −1-1⁠.
    Formula (i), Section 7.2; Property V, Section 7.2.1. Basics: the power rule

  4. Check: differentiate the answer back

    Let F(x)=67x7/2+45x5/2+2x3/2+CF(x) = \frac67 x^{7/2} + \frac45 x^{5/2} + 2x^{3/2} + C⁠. Then

    F′(x)=67⋅72x5/2+45⋅52x3/2+2⋅32x1/2=3x5/2+2x3/2+3x1/2=x (3x2+2x+3).\begin{aligned} F'(x) &= \tfrac67\cdot\tfrac72 x^{5/2} + \tfrac45\cdot\tfrac52 x^{3/2} \\ &\quad + 2\cdot\tfrac32 x^{1/2} \\ &= 3x^{5/2} + 2x^{3/2} + 3x^{1/2} \\ &= \sqrt{x}\,(3x^2 + 2x + 3). \end{aligned}

    This is the integrand, so the answer is correct.

    67⋅72=3\frac67\cdot\frac72 = 3⁠, 45⋅52=2\frac45\cdot\frac52 = 2⁠, 2⋅32=32\cdot\frac32 = 3⁠: each coefficient returns to the original one. The last line takes out the common factor x1/2x^{1/2}⁠, reversing Step 2.
    Power rule for derivatives, ddxxm=mxm−1\frac{d}{dx}x^m = m x^{m-1}⁠; Property I, Section 7.2.1. Basics: integration undoes differentiation
Result

For x≥0x \ge 0⁠,

∫x (3x2+2x+3) dx=67x7/2+45x5/2+2x3/2+C.\begin{aligned} &\int \sqrt{x}\,(3x^2+2x+3)\,dx \\ &\quad = \frac67 x^{7/2} + \frac45 x^{5/2} + 2x^{3/2} + C. \end{aligned}

(Textbook answer: 67x7/2+45x5/2+2x3/2+C\frac67 x^{7/2} + \frac45 x^{5/2} + 2x^{3/2} + C⁠.)

Check it with numbers

At x=4x = 4 the integrand is 2⋅(48+8+3)=1182\cdot(48 + 8 + 3) = 118⁠. With F(x)=67x7/2+45x5/2+2x3/2F(x) = \frac67 x^{7/2} + \frac45 x^{5/2} + 2x^{3/2}⁠: F(4.001)≈151.43231909F(4.001) \approx 151.43231909 and F(3.999)≈151.19631909F(3.999) \approx 151.19631909⁠, so the slope of FF at 4 is about

151.43231909−151.196319090.002≈118.0000.\frac{151.43231909 - 151.19631909}{0.002} \approx 118.0000.

It matches 118.

(Each value of FF is rounded to 8 decimal places, so the slope is reliable to about 4 decimal places.)

Common mistakes
  1. Multiplying x\sqrt{x} into only the first term. All three terms of the bracket get the factor x1/2x^{1/2}⁠, including the constant 3.
  2. Losing the constants. ∫3x5/2 dx=3⋅27x7/2=67x7/2\int 3x^{5/2}\,dx = 3\cdot\frac27 x^{7/2} = \frac67 x^{7/2}⁠, not 27x7/2\frac27 x^{7/2}⁠.
  3. Using the old exponent in the denominator. Divide by the new power 72\frac72⁠, not by 52\frac52⁠.
What to take away
  1. Expand products into a sum of constant multiples of powers, then integrate each.
  2. For ∫k xn dx\int k\,x^n\,dx with n≠−1n \ne -1⁠, the coefficient of the answer is kn+1\frac{k}{n+1}⁠.
  3. Differentiating back recovers every coefficient, so it catches arithmetic slips.

Every fact used, and where it comes from

FactUsed inSource
Keep the interval of definition in mindStep 1Note after the list of formulae, Section 7.2
xmxn=xm+nx^m x^n = x^{m+n}⁠; distributive lawStep 2Laws of exponents and algebra (earlier classes)
∫xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C⁠, n≠−1n \ne -1Step 3Formula (i), Section 7.2
Integrate term by term, constants in frontStep 3Property V, Section 7.2.1 (Lesson: integration as anti-differentiation)
ddxxm=mxm−1\frac{d}{dx}x^m = m x^{m-1}⁠; ddx∫f dx=f\frac{d}{dx}\int f\,dx = fStep 4Derivatives column of formula (i), Section 7.2; Property I, Section 7.2.1
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