Integrals: practice problems with step-by-step solutions

Free Integrals practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: polynomials, exponentials, logs and substitution, areas between curves.

Free practice problemEasy
xy−2−1

Evaluate I = ∫−2−1 (4x2 − 4x + 5) dx.

f(x) = 4x2 − 4x + 5[−2, −1]
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What you practise

  1. Easypolynomials
  2. Mediumexponentials, logs and substitution
  3. Hardareas between curves

How to solve Integrals problems

  1. Rewrite the integrand as a sum of simple terms where possible.
  2. Apply the power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1), and standard integrals for ex, 1/x, sin and cos.
  3. Use substitution when you see a function and its derivative together.
  4. For a definite integral, evaluate F(b) − F(a).

Common mistakes

  • Forgetting to divide by the new power.
  • Not changing the limits after a substitution.
  • Forgetting + C in an indefinite integral.

Key formulas

Power rule
∫xⁿ dx = xⁿ⁺¹/(n + 1) + C
Definite integral
∫ab f dx = F(b) − F(a)
Common integrals
∫eˣ dx = eˣ, ∫1/x dx = ln|x|, ∫sin x dx = −cos x
Area between curves
∫ab (f − g) dx

Learn Integrals from scratch

Before this: Derivatives

1What an integral is

An integral adds up a quantity that changes. The clearest picture is an area under a graph.

A car drives at 20 m/s for 5 s. It covers 20 · 5 = 100 m. On a speed–time graph, that is the area of the rectangle under the line.

xy520100

When the speed changes, the shape under the graph is curved, but the idea is the same: the area is the distance.

We write the area under f between x = a and x = b as:

∫ab f(x) dx

The numbers a and b are the limits of the integral. The dx tells you that x is the variable.

Self-check A cyclist rides at a constant 10 m/s for 4 s. What is the area under the speed–time graph?
  1. 10 m
  2. 14 m
  3. 2.5 m
  4. 40 m

Answer: 40 m. The area is a rectangle: height 10 m/s times width 4 s gives 40 m. Dividing 10 by 4 gives 2.5, which is not an area.

2Antiderivatives and the power rule

Integration is differentiation backwards. An antiderivative F of f is a function whose derivative is f.

Since (x³)′ = 3x², an antiderivative of 3x² is x³. Any constant could be added, so we write ∫3x² dx = x³ + C.

The power rule for integrals: raise the power by one, then divide by the new power.

∫xⁿ dx = xⁿ⁺¹/(n + 1) + C, n ≠ −1
  • A constant factor stays in front: ∫5x² dx = 5x³/3 + C.
  • A constant on its own: ∫4 dx = 4x + C.
  • In a sum, integrate every term separately.

Learn these too: ∫eˣ dx = eˣ, ∫ekx dx = ekx/k, ∫1/x dx = ln|x|, ∫cos x dx = sin x and ∫sin x dx = −cos x.

You can always check an antiderivative: differentiate it, and you must get the original function back.

Self-check What is ∫6x² dx?
  1. 12x + C
  2. 2x³ + C
  3. 3x³ + C
  4. 6x³ + C

Answer: 2x³ + C. Raise the power to 3 and divide by 3: 6x³/3 = 2x³. 12x is the derivative, not the integral, and 6x³ forgets to divide by the new power.

3Definite integrals and areas

A definite integral has limits and gives a number. Find an antiderivative F, then subtract its values at the two limits.

∫ab f(x) dx = F(b) − F(a)

The constant C cancels in the subtraction, so you can leave it out here.

Example: ∫13 3/x dx = 3 ln 3 − 3 ln 1 = 3 ln 3 ≈ 3.296, because ln 1 = 0.

Substitution helps when a function sits inside another one. For e0.5x, set u = 0.5x. The result is the rule above: divide by the 0.5.

∫03 e0.5x dx = 2e0.5x |03 = 2(e1.5 − 1) ≈ 6.963

The area between two curves is the integral of the top curve minus the bottom curve, between the points where they meet.

For y = 2x and y = x², the curves meet at x = 0 and x = 2. On that interval 2x is on top.

A = ∫02 (2x − x²) dx = 4 − 8/3 = 4/3
Self-check The curves y = x and y = x² meet at x = 0 and x = 1. Between them, y = x is on top. What is the enclosed area?
  1. 1/2
  2. 1/6
  3. 5/6
  4. −1/6

Answer: 1/6. ∫(x − x²) dx from 0 to 1 is 1/2 − 1/3 = 1/6. Taking bottom minus top gives −1/6, and adding the two integrals gives 5/6.

4The method step by step

  1. Rewrite the function as a sum of simple terms. Write roots and fractions as powers: √x = x1/2.
  2. Find an antiderivative of every term with the power rule or a standard integral.
  3. Check it quickly by differentiating in your head.
  4. For a definite integral, work out F(b) and F(a) separately.
  5. Subtract: F(b) − F(a). Watch the minus signs when a is negative.

For an area between curves, first solve f(x) = g(x) to find the limits, then integrate top minus bottom.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

Evaluate I = ∫02 (3x² + 2x + 1) dx.

1

Integrate term by term

3x² gives 3x³/3 = x³. 2x gives 2x²/2 = x². The constant 1 gives x.

F(x) = x³ + x² + x
2

Check by differentiating

F′(x) = 3x² + 2x + 1

This is the original function, so F is correct.

3

Value at the upper limit

F(2) = 8 + 4 + 2 = 14
4

Value at the lower limit

F(0) = 0 + 0 + 0 = 0
5

Subtract

I = F(2) − F(0) = 14 − 0 = 14
xy214

The integral is the shaded area under the curve between x = 0 and x = 2.

Answer

I = 14

6Common mistakes

  • Forgetting to divide by the new power.
  • Differentiating instead of integrating: lowering the power instead of raising it.
  • Writing ∫1/x dx as x⁰/0. The power rule does not work for n = −1; use ln|x|.
  • Forgetting the minus signs in F(a) when a is negative, for example (−1)³ = −1.
  • Forgetting + C in an indefinite integral, or not changing the limits after a substitution.
Self-check What is ∫02 x³ dx?
  1. 12
  2. 16
  3. 4
  4. 8

Answer: 4. F(x) = x⁴/4, so F(2) − F(0) = 16/4 − 0 = 4. Forgetting to divide by the new power 4 gives 16.

7How exam questions look

  • Easy: a definite integral of a polynomial, sometimes with negative limits. Use the power rule and F(b) − F(a).
  • Medium: exponentials, logs and substitution, such as ∫e0.5x dx or ∫3/x dx between two limits.
  • Hard: the area enclosed by two curves, such as y = x² and y = 4x. Find where they meet, then integrate top minus bottom.

A quick check: an area enclosed by two curves is always positive. If you get a negative number, you subtracted in the wrong order.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
xy3y = 0.5x

The region under y = 0.5x from x = 0 to x = 3 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = 0.5x0 ≤ x ≤ 3

Find V

Show hints
  1. A slice at x is a disc of radius y = 0.5x.
  2. V = π∫₀3 (0.5x)² dx.
  3. = π·0.25·x³/3.
Show the answer and the step-by-step solution

Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.

Disc method

V = π∫₀3 0.25x² dx

Integrate

V = π·0.25·3³/3 = 7.069

Answer

V = 7.069

Check: a cone of radius 1.5 and height 3 has (1/3)πr²h = 7.069.

Example 2Medium
xy2y = x·cos x

Evaluate I = ∫₀2 x·cos x dx (x in radians).

∫₀2 x·cos x dx

Find I

Show hints
  1. Take u = x and dv = cos x dx.
  2. v = sin x.
  3. ∫x cos x dx = x sin x + cos x.
Show the answer and the step-by-step solution

Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.

Choose u and dv

u = x, du = dx; dv = cos x dx, v = sin x

Antiderivative

∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x

Limits

(2·sin 2 + cos 2) − (0 + 1) = 0.4024

Answer

I = 0.4024
Example 3Hard
xyy = 2x2​y = x

Find the area A of the region enclosed by y = 2x2 and y = x.

y = 2x2y = x

Find A

Show hints
  1. First find where the curves meet: set them equal.
  2. They meet at x = 0 and x = 0.5. Between these the line lies above the parabola.
  3. A = ∫00.5 (x − 2x2) dx.
Show the answer and the step-by-step solution

Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.

Intersection points

2x2 = x → x = 0, x = 0.5

Integrate upper minus lower

A = [0.5x2 − 0.6667x3]00.5 = 0.04167

Answer

A = 0.04167

Questions and answers

What is Integrals?

An integral adds up a quantity that changes. The clearest picture is an area under a graph.

A car drives at 20 m/s for 5 s. It covers 20 · 5 = 100 m. On a speed–time graph, that is the area of the rectangle under the line.

How do I solve Integrals problems step by step?
  1. Rewrite the integrand as a sum of simple terms where possible.
  2. Apply the power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1), and standard integrals for ex, 1/x, sin and cos.
  3. Use substitution when you see a function and its derivative together.
  4. For a definite integral, evaluate F(b) − F(a).
What are common mistakes in Integrals?
  • Forgetting to divide by the new power.
  • Not changing the limits after a substitution.
  • Forgetting + C in an indefinite integral.
What do the Integrals exercises cover?
  • Easy: polynomials
  • Medium: exponentials, logs and substitution
  • Hard: areas between curves

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

Derivatives

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Convolution & LTI systemsLaplace transformsMultivariable calculusNumerical methodsDifferential equations

Calculators

Integral calculator (definite and indefinite)

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Printable worksheet

Integrals worksheet with answers
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