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Chapter 7 · Integrals

Exercise 7.1 · Q17: ∫(2x2−3sin⁡x+5x) dx\int (2x^2 - 3\sin x + 5\sqrt{x})\,dx

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

Find the following integral:

∫(2x2−3sin⁡x+5x) dx\int (2x^2 - 3\sin x + 5\sqrt{x})\,dx
Exam answer

Let I=∫(2x2−3sin⁡x+5x)dxI = \displaystyle\int\left(2x^2 - 3\sin x + 5\sqrt{x}\right)dx⁠.

I=2∫x2 dx−3∫sin⁡x dx+5∫x1/2 dx=2x33−3(−cos⁡x)+5⋅x3/23/2+C\begin{aligned} I &= 2\int x^2\,dx - 3\int\sin x\,dx \\ &\quad + 5\int x^{1/2}\,dx \\ &= \frac{2x^3}{3} - 3(-\cos x) \\ &\quad + 5\cdot\frac{x^{3/2}}{3/2} + C \end{aligned}

∴ I=23x3+3cos⁡x\therefore\ I = \dfrac23x^3 + 3\cos x +103x3/2+C+ \dfrac{10}{3}x^{3/2} + C⁠.

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

The one idea this problem is built on

Split the integral into three standard ones (Property V). The only delicate point is a double minus sign: ∫sin⁡x dx=−cos⁡x\int \sin x\,dx = -\cos x⁠, so −3∫sin⁡x dx=−3(−cos⁡x)=+3cos⁡x-3\int\sin x\,dx = -3(-\cos x) = +3\cos x⁠. The term 5x=5x1/25\sqrt{x} = 5x^{1/2} is a power of xx and goes through the power formula.

Given, to find, and the plan
Given
The integrand f(x)=2x2−3sin⁡x+5xf(x) = 2x^2 - 3\sin x + 5\sqrt{x}⁠.
To find
∫f(x) dx\int f(x)\,dx⁠.
Plan
(1) Note where ff is defined. (2) Split into three integrals, constants in front. (3) Use the standard formulae for xnx^n and sin⁡x\sin x⁠. (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

2The integral of sin⁡x\sin x

ddx(−cos⁡x)=sin⁡x\frac{d}{dx}(-\cos x) = \sin x⁠, so ∫sin⁡x dx=−cos⁡x+C\int \sin x\,dx = -\cos x + C (formula (iii), Section 7.2). Basics: trigonometric facts for integrals

3The power formula and roots

∫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), with x=x1/2\sqrt{x} = x^{1/2}⁠. Basics: the power rule

4Linearity

∫(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?

    2x22x^2 and sin⁡x\sin x are defined for all real xx⁠; x\sqrt{x} needs x≥0x \ge 0⁠. So the integrand is defined for x≥0x \ge 0⁠.

    The answer is valid on [0,∞)[0, \infty)⁠, with the check at the end point x=0x = 0 holding as a right-hand derivative.
    The Note in Section 7.2: always keep the interval of definition in mind.

  2. Split the integral (Property V)

    ∫(2x2−3sin⁡x+5x) dx=2 ⁣∫x2 dx−3 ⁣∫sin⁡x dx+5 ⁣∫x1/2 dx.\begin{aligned} &\int (2x^2 - 3\sin x + 5\sqrt{x})\,dx \\ &\quad = 2\!\int x^2\,dx - 3\!\int \sin x\,dx + 5\!\int x^{1/2}\,dx. \end{aligned}
    Property V with k1=2k_1 = 2⁠, k2=−3k_2 = -3⁠, k3=5k_3 = 5⁠. Writing x\sqrt{x} as x1/2x^{1/2} makes the last integral a power of xx⁠. We keep the minus sign with the 3, since it will meet another minus sign in the next step.
    Property V, Section 7.2.1. Lesson: integration as anti-differentiation

  3. Integrate each piece

    2∫x2 dx=2⋅x33=23x3,−3∫sin⁡x dx=−3(−cos⁡x)=3cos⁡x,5∫x1/2 dx=5⋅x3/23/2=103x3/2.\begin{aligned} 2\int x^2\,dx &= 2\cdot\frac{x^3}{3} = \frac23 x^3, \\ -3\int \sin x\,dx &= -3(-\cos x) = 3\cos x, \\ 5\int x^{1/2}\,dx &= 5\cdot\frac{x^{3/2}}{3/2} = \frac{10}{3}x^{3/2}. \end{aligned}

    Adding, with a single constant,

    ∫(2x2−3sin⁡x+5x) dx=23x3+3cos⁡x+103x3/2+C.\begin{aligned} &\int (2x^2 - 3\sin x + 5\sqrt{x})\,dx \\ &\quad = \frac23 x^3 + 3\cos x + \frac{10}{3}x^{3/2} + C. \end{aligned}
    Power formula with n=2n = 2 gives x33\frac{x^3}{3}⁠. The sine integral is −cos⁡x-\cos x because the derivative of cos⁡x\cos x is −sin⁡x-\sin x⁠; multiplying by −3-3 turns it into +3cos⁡x+3\cos x⁠. For n=12n = \frac12 the new power is 32\frac32⁠, and dividing by 32\frac32 means multiplying by 23\frac23⁠: 5⋅23=1035\cdot\frac23 = \frac{10}{3}⁠.
    Formulae (i) and (iii), Section 7.2. Basics: trigonometric facts for integrals

  4. Check: differentiate the answer back

    Let F(x)=23x3+3cos⁡x+103x3/2+CF(x) = \frac23 x^3 + 3\cos x + \frac{10}{3}x^{3/2} + C⁠. Then

    F′(x)=23⋅3x2+3(−sin⁡x)+103⋅32x1/2=2x2−3sin⁡x+5x.\begin{aligned} F'(x) &= \tfrac23\cdot 3x^2 + 3(-\sin x) \\ &\quad + \tfrac{10}{3}\cdot\tfrac32 x^{1/2} \\ &= 2x^2 - 3\sin x + 5\sqrt{x}. \end{aligned}

    This is the integrand, so the answer is correct.

    ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin x brings the minus sign back, and 103⋅32=5\frac{10}{3}\cdot\frac32 = 5 recovers the coefficient of x\sqrt{x}⁠. Every term returns exactly, which also confirms the sign of 3cos⁡x3\cos x⁠.
    Derivatives column of the list in Section 7.2; Property I, Section 7.2.1. Basics: integration undoes differentiation
Result

For x≥0x \ge 0⁠,

∫(2x2−3sin⁡x+5x) dx=23x3+3cos⁡x+103x3/2+C.\begin{aligned} &\int (2x^2 - 3\sin x + 5\sqrt{x})\,dx \\ &\quad = \frac23 x^3 + 3\cos x + \frac{10}{3}x^{3/2} + C. \end{aligned}

(Textbook answer: 23x3+3cos⁡x+103x3/2+C\frac23 x^3 + 3\cos x + \frac{10}{3}x^{3/2} + C⁠.)

Check it with numbers

At x=1x = 1 (radians) the integrand is 2−3sin⁡1+52 - 3\sin 1 + 5 ≈7−2.5244130\approx 7 - 2.5244130 =4.4755870= 4.4755870⁠. With F(x)=23x3+3cos⁡x+103x3/2F(x) = \frac23 x^3 + 3\cos x + \frac{10}{3}x^{3/2}⁠: F(1.001)≈5.62538495F(1.001) \approx 5.62538495 and F(0.999)≈5.61643377F(0.999) \approx 5.61643377⁠, so the slope of FF at 1 is about

5.62538495−5.616433770.002≈4.4756,\frac{5.62538495 - 5.61643377}{0.002} \approx 4.4756,

which matches 4.4755870 to four decimal places. Had we written −3cos⁡x-3\cos x instead, the slope at 1 would be about 9.524, far from the integrand.

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

Common mistakes
  1. The double minus. −3∫sin⁡x dx=−3(−cos⁡x)=+3cos⁡x-3\int\sin x\,dx = -3(-\cos x) = +3\cos x⁠. Writing −3cos⁡x-3\cos x is the most common slip here.
  2. ∫x dx=12x\int\sqrt{x}\,dx = \frac{1}{2\sqrt{x}}⁠. That is the derivative of x\sqrt{x}⁠. The integral is 23x3/2\frac23 x^{3/2}⁠.
  3. Ignoring the domain. The x\sqrt{x} term restricts the answer to x≥0x \ge 0⁠.
What to take away
  1. ∫sin⁡x dx=−cos⁡x+C\int\sin x\,dx = -\cos x + C and ∫cos⁡x dx=sin⁡x+C\int\cos x\,dx = \sin x + C⁠: only the sine integral carries a minus sign.
  2. Keep each constant together with its sign when splitting an integral.
  3. Differentiating back is the quickest way to catch a sign error in a trigonometric term.

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
Integrate term by term, constants in frontStep 2Property V, Section 7.2.1 (Lesson: integration as anti-differentiation)
∫xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C⁠; x=x1/2\sqrt{x} = x^{1/2}Step 3Formula (i), Section 7.2; laws of exponents
∫sin⁡x dx=−cos⁡x+C\int \sin x\,dx = -\cos x + CStep 3Formula (iii), Section 7.2
(cos⁡x)′=−sin⁡x(\cos x)' = -\sin x⁠; ddxxm=mxm−1\frac{d}{dx}x^m = m x^{m-1}Step 4Derivatives column, list in Section 7.2; Property I, Section 7.2.1
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