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Chapter 7 · Integrals

Exercise 7.1 · Q14: ∫(1−x)x dx\int (1-x)\sqrt{x}\,dx

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

Find the following integral:

∫(1−x)x dx\int (1 - x)\sqrt{x}\,dx
Exam answer

Let I=∫(1−x)x dxI = \displaystyle\int(1 - x)\sqrt{x}\,dx⁠.

I=∫(x1/2−x3/2)dx=x3/23/2−x5/25/2+C\begin{aligned} I &= \int\left(x^{1/2} - x^{3/2}\right)dx \\ &= \frac{x^{3/2}}{3/2} - \frac{x^{5/2}}{5/2} + C \end{aligned}

∴ I=23x3/2−25x5/2+C\therefore\ I = \dfrac23x^{3/2} - \dfrac25x^{5/2} + C⁠.

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

The one idea this problem is built on

There is no product rule for integrals, so we cannot integrate (1−x)(1 - x) and x\sqrt{x} separately and multiply. Instead we multiply out first: (1−x)x=x1/2−x3/2(1 - x)\sqrt{x} = x^{1/2} - x^{3/2}⁠. Now the integrand is a difference of two powers of xx⁠, and each is integrated by the power formula.

Given, to find, and the plan
Given
The integrand f(x)=(1−x)xf(x) = (1 - x)\sqrt{x}⁠.
To find
∫f(x) dx\int f(x)\,dx⁠.
Plan
(1) Note where ff is defined. (2) Expand the product into powers of xx⁠. (3) Integrate term by term. (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

x=x1/2\sqrt{x} = x^{1/2} and xm⋅xn=xm+nx^m\cdot x^n = x^{m+n}⁠. So x⋅x=x1⋅x1/2=x3/2x\cdot\sqrt{x} = x^{1}\cdot x^{1/2} = 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

4Differences and constant multiples

∫(f−g) dx=∫f dx−∫g dx\int (f - g)\,dx = \int f\,dx - \int g\,dx (Property V, Section 7.2.1, with k1=1k_1 = 1⁠, k2=−1k_2 = -1⁠). Lesson: integration as anti-differentiation

Solution

  1. Where is the integrand defined?

    x\sqrt{x} is real only for x≥0x \ge 0⁠, and nothing else restricts xx⁠. So the integrand lives on

    x≥0.x \ge 0.
    The answer is valid where the integrand is defined. At the end point x=0x = 0 a derivative can only be taken from the right, so the check in Step 4 is for x>0x > 0⁠, with x=0x = 0 included as a right-hand derivative.
    The Note in Section 7.2: always keep the interval of definition in mind.

  2. Expand the product into powers of xx

    (1−x)x=1⋅x1/2−x⋅x1/2=x1/2−x1+12=x1/2−x3/2.\begin{aligned} (1 - x)\sqrt{x} &= 1\cdot x^{1/2} - x\cdot x^{1/2} \\ &= x^{1/2} - x^{1 + \frac12} \\ &= x^{1/2} - x^{3/2}. \end{aligned}
    The integral of a product is not the product of the integrals, so the product must go. Multiplying out (the distributive law) turns it into a difference of two single powers, which the power formula handles. The exponents add because the bases are equal: 1+12=321 + \frac12 = \frac32⁠.
    xm⋅xn=xm+nx^m\cdot x^n = x^{m+n} for x>0x > 0⁠. Basics: powers of x

  3. Integrate term by term

    ∫x1/2 dx=x3/23/2=23x3/2,∫x3/2 dx=x5/25/2=25x5/2.\begin{aligned} \int x^{1/2}\,dx &= \frac{x^{3/2}}{3/2} = \frac23 x^{3/2}, \\ \int x^{3/2}\,dx &= \frac{x^{5/2}}{5/2} = \frac25 x^{5/2}. \end{aligned}

    So

    ∫(1−x)x dx=23x3/2−25x5/2+C.\int (1 - x)\sqrt{x}\,dx = \frac23 x^{3/2} - \frac25 x^{5/2} + C.
    Power formula with n=12n = \frac12 (new power 32\frac32⁠) and n=32n = \frac32 (new power 52\frac52⁠). Dividing by 32\frac32 is multiplying by 23\frac23⁠; dividing by 52\frac52 is multiplying by 25\frac25⁠. The minus sign between the terms is carried through unchanged (Property V), and the two constants combine into one CC⁠.
    Formula (i), Section 7.2; Property V, Section 7.2.1. Basics: the power rule

  4. Check: differentiate the answer back

    Let F(x)=23x3/2−25x5/2+CF(x) = \frac23 x^{3/2} - \frac25 x^{5/2} + C⁠. Then

    F′(x)=23⋅32x1/2−25⋅52x3/2=x1/2−x3/2=(1−x)x.\begin{aligned} F'(x) &= \tfrac23\cdot\tfrac32 x^{1/2} - \tfrac25\cdot\tfrac52 x^{3/2} \\ &= x^{1/2} - x^{3/2} \\ &= (1 - x)\sqrt{x}. \end{aligned}

    This is the integrand, so the answer is correct.

    Each coefficient was chosen in Step 3 precisely so that it cancels the factor brought down by differentiation: 23⋅32=1\frac23\cdot\frac32 = 1 and 25⋅52=1\frac25\cdot\frac52 = 1⁠. The last line is Step 2 read backwards (take out the common factor x1/2x^{1/2}⁠).
    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,

∫(1−x)x dx=23x3/2−25x5/2+C.\int (1 - x)\sqrt{x}\,dx = \frac23 x^{3/2} - \frac25 x^{5/2} + C.

(Textbook answer: 23x3/2−25x5/2+C\frac23 x^{3/2} - \frac25 x^{5/2} + C⁠.)

Check it with numbers

At x=4x = 4 the integrand is (1−4)⋅2=−6(1 - 4)\cdot 2 = -6⁠. With F(x)=23x3/2−25x5/2F(x) = \frac23 x^{3/2} - \frac25 x^{5/2}⁠: F(4.001)≈−7.47266804F(4.001) \approx -7.47266804 and F(3.999)≈−7.46066804F(3.999) \approx -7.46066804⁠, so the slope of FF at 4 is about

−7.47266804−(−7.46066804)0.002≈−6.0000.\frac{-7.47266804 - (-7.46066804)}{0.002} \approx -6.0000.

It matches −6-6⁠. The negative slope also makes sense: for x>1x > 1 the factor 1−x1 - x is negative, so FF is decreasing there.

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

Common mistakes
  1. Integrating the factors separately. ∫(1−x)x dx≠∫(1−x) dx⋅∫x dx\int (1-x)\sqrt{x}\,dx \ne \int (1-x)\,dx\cdot\int \sqrt{x}\,dx⁠. Expand first.
  2. Multiplying exponents instead of adding. x⋅x1/2=x3/2x\cdot x^{1/2} = x^{3/2}⁠, not x1/2x^{1/2}⁠.
  3. Coefficient upside down. ∫x3/2dx=25x5/2\int x^{3/2}dx = \frac25 x^{5/2}⁠, not 52x5/2\frac52 x^{5/2}⁠.
What to take away
  1. Products of simple factors are expanded into a sum of powers before integrating.
  2. For ∫xp/q dx\int x^{p/q}\,dx the answer is qp+qx(p+q)/q+C\frac{q}{p+q}x^{(p+q)/q} + C (when p+q≠0p + q \ne 0⁠): raise the power, multiply by the reciprocal of the new power.
  3. Differentiating back checks every coefficient at once.

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
x=x1/2\sqrt{x} = x^{1/2}⁠; 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
∫(k1f1+k2f2) dx=k1∫f1+k2∫f2\int (k_1 f_1 + k_2 f_2)\,dx = k_1\int f_1 + k_2\int f_2Step 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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