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Grade 12 · Lesson 4.1

Antiderivatives

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Goals

Remember

Exercises

Exercise 1

Find ∫(4x3−6x+7) dx\displaystyle\int(4x^3-6x+7)\,dx.

Exercise 2

Find ∫(3x+1x2)dx\displaystyle\int\left(3\sqrt x+\frac1{x^2}\right)dx on (0,+∞)(0, +\infty).

Exercise 3

Find ∫(5ex−2sin⁡x+1sin⁡2x)dx\displaystyle\int\left(5e^x-2\sin x+\frac{1}{\sin^2x}\right)dx.

Exercise 4

Find ∫(2x−1)2 dx\displaystyle\int(2x-1)^2\,dx.

Exercise 5

Consider f(x)=3x2+2cos⁡xf(x)=3x^2+2\cos x. Is each claim true or false?

a) F(x)=x3+2sin⁡xF(x)=x^3+2\sin x is an antiderivative of f(x)f(x).

b) F(x)=x3−2sin⁡x+5F(x)=x^3-2\sin x+5 is an antiderivative of f(x)f(x).

c) The antiderivative G(x)G(x) of f(x)f(x) with G(0)=1G(0)=1 is G(x)=x3+2sin⁡x+1G(x)=x^3+2\sin x+1.

d) For G(x)G(x) in c), G(π2)=π38+2G\left(\dfrac{\pi}{2}\right)=\dfrac{\pi^3}{8}+2.

Exercise 6

F(x)F(x) is the antiderivative of f(x)=4x3−2xf(x)=4x^3-2x with F(1)=3F(1)=3. Find F(2)F(2).

Exercise 7

A tank already holds 200200 litres of water. A pump is switched on, and tt minutes later the volume of water is changing at the rate V′(t)=6t+20V'(t)=6t+20 (litres per minute). How many litres are in the tank after 1010 minutes?

Exercise 8 Review

An object moves in a straight line with velocity v(t)=12t−3t2v(t)=12t-3t^2 (m/s), 0≤t≤40\le t\le 4. Find its greatest velocity and the distance it travels in the first 44 seconds.

Exercise 9 Challenge

Find the function f(x)f(x) with f′′(x)=6x+2f''(x)=6x+2, f′(0)=1f'(0)=1 and f(0)=−1f(0)=-1.

Exercise 10 Challenge

Find ∫sin⁡2x2 dx\displaystyle\int\sin^2\frac{x}{2}\,dx.

Mini test

15 minutes, without looking back at the lesson. Part I: circle one letter; Part II: write T or F for each statement; Part III: write the answer in the box.

Part I. Multiple choice

Question 1. The indefinite integral of f(x)=3x2−sin⁡xf(x)=3x^2-\sin x is
A. x3+cos⁡x+Cx^3+\cos x+C.B. x3−cos⁡x+Cx^3-\cos x+C.C. 6x−cos⁡x+C6x-\cos x+C.D. 3x3+cos⁡x+C3x^3+\cos x+C.
Question 2. The indefinite integral of f(x)=2xf(x)=2^x is
A. 2xln⁡2+C2^x\ln 2+C.B. 2xln⁡2+C\dfrac{2^x}{\ln 2}+C.C. 2x+1x+1+C\dfrac{2^{x+1}}{x+1}+C.D. x⋅2x−1+Cx\cdot 2^{x-1}+C.
Question 3. F(x)F(x) is the antiderivative of f(x)=2x+1f(x)=2x+1 with F(1)=5F(1)=5. The value of F(2)F(2) is
A. 66.B. 1111.C. 99.D. 55.
Question 4. A car brakes with velocity v(t)=15−3tv(t)=15-3t (m/s) until it stops. The distance it travels from braking until it stops is
A. 7575 m.B. 37.537.5 m.C. 112.5112.5 m.D. 00 m.

Part II. True or false

Question 1. Consider f(x)=ex−2xf(x)=e^x-2x.
a) F(x)=ex−x2F(x)=e^x-x^2 is an antiderivative of f(x)f(x).T / F
b) ∫f(x) dx=ex−2+C\displaystyle\int f(x)\,dx=e^x-2+C.T / F
c) The antiderivative F(x)F(x) of f(x)f(x) with F(0)=3F(0)=3 is F(x)=ex−x2+2F(x)=e^x-x^2+2.T / F
d) For F(x)F(x) in c), F(1)=e+1F(1)=e+1. T / F

Part III. Short answer

Question 1. F(x)F(x) is the antiderivative of f(x)=6x2−4x+3f(x)=6x^2-4x+3 with F(0)=−2F(0)=-2. Find F(2)F(2).
Question 2. F(x)F(x) is the antiderivative of f(x)=ex+1xf(x)=e^x+\dfrac1x on (0,+∞)(0, +\infty) with F(1)=eF(1)=e. Find F(2)F(2). Give your answer to 1 decimal place.

Score: ........ / 10

Hints (only when stuck)

Exercise 1.

What you need: ∫xαdx=xα+1α+1+C\displaystyle\int x^\alpha dx=\frac{x^{\alpha+1}}{\alpha+1}+C and the rule for sums.

Steps:

  1. Find the antiderivative of each term.
  2. Add a constant CC and check by differentiating.
Exercise 2.

What you need: write x=x12\sqrt x=x^{\frac12} and 1x2=x−2\dfrac{1}{x^2}=x^{-2}, then use the power formula.

Steps:

  1. Rewrite as powers.
  2. Apply the formula to each term.
Exercise 3.

What you need: the exe^x, sin⁡x\sin x and 1sin⁡2x\dfrac{1}{\sin^2x} rows of the table.

Steps:

  1. Write the antiderivative of each term, watching the two minus signs.
  2. Put them together and add CC.
Exercise 4.

What you need: do not find the antiderivative of a square by "squaring the antiderivative"; expand first.

Steps:

  1. Expand (2x−1)2=4x2−4x+1(2x-1)^2=4x^2-4x+1.
  2. Find the antiderivative of the polynomial.
Exercise 5.

What you need: FF is an antiderivative of ff when F′=fF'=f; the condition G(0)=1G(0)=1 fixes the constant CC.

Steps:

  1. Differentiate the functions in a) and b).
  2. Find CC in c), then substitute x=π2x=\dfrac{\pi}{2} for d).
Exercise 6.

What you need: find the family of antiderivatives, then use F(1)=3F(1)=3 to find CC.

Steps:

  1. F(x)=x4−x2+CF(x)=x^4-x^2+C.
  2. Substitute x=1x=1 to find CC, then compute F(2)F(2).
Exercise 7.

What you need: the volume V(t)V(t) is an antiderivative of V′(t)V'(t); at the start, V(0)=200V(0)=200.

Steps:

  1. Find V(t)=3t2+20t+CV(t)=3t^2+20t+C.
  2. Use V(0)=200V(0)=200 to find CC, then compute V(10)V(10).
Exercise 8.

What you need: how to find a maximum on a closed interval (see maximum and minimum values); distance is an antiderivative of velocity.

Steps:

  1. Solve v′(t)=0v'(t)=0 and compare v(0)v(0), v(2)v(2), v(4)v(4).
  2. Find s(t)s(t) with s(0)=0s(0)=0, then compute s(4)s(4).
Exercise 9.

What you need: f′f' is an antiderivative of f′′f'', and ff is an antiderivative of f′f' (second derivatives were met in grade 11).

Steps:

  1. Find f′(x)f'(x) and use f′(0)=1f'(0)=1 to find the first constant.
  2. Find f(x)f(x) and use f(0)=−1f(0)=-1 to find the second constant.
Exercise 10.

What you need: the double-angle formula (grade 11): cos⁡x=1−2sin⁡2x2\cos x=1-2\sin^2\dfrac x2, so sin⁡2x2=1−cos⁡x2\sin^2\dfrac x2=\dfrac{1-\cos x}{2}.

Steps:

  1. Replace sin⁡2x2\sin^2\dfrac x2 by 12−12cos⁡x\dfrac12-\dfrac12\cos x.
  2. Find the antiderivative term by term.

Self-check

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