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Calculus I · Applications of Integrals

Area Between Curves — Practice Problems and Worked Solutions

Integrate top minus bottom across the interval.

The formulas you need here

Area between two curves

A = \int_{a}^{b} [f (x) - g (x)] d x

What it says: Upper curve minus lower curve, added up across the interval.
When to reach for it: The region trapped between two graphs.
Watch out: Find where they cross first — those are your limits, and the upper curve may swap on the other side of a crossing.

A = \int_{0}^{1} (x - x^{2}) d x = \frac{1}{6}

On [0,1] the line sits above the parabola throughout.

16 worked problems

Every step is generated and checked by a computer algebra system.

1.
Find the area between the curves.
$y = x, y = x^{2}, 0 \leq x \leq 1$
Answer: $\frac{1}{6}$
Step-by-step solution
1. $A = \int_{a}^{b} (top - bottom) d x$
2. $A = \int_{0}^{1} (x - (x^{2})) d x = \frac{1}{6}$
2.
Find the area between the curves.
$y = 2 x, y = x^{2}, 0 \leq x \leq 2$
Answer: $\frac{4}{3}$
Step-by-step solution
1. $A = \int_{a}^{b} (top - bottom) d x$
2. $A = \int_{0}^{2} (2 x - (x^{2})) d x = \frac{4}{3}$
3.
Find the area between the curves.
$y = 4 - x^{2}, y = 0, - 2 \leq x \leq 2$
Answer: $\frac{32}{3}$
Step-by-step solution
1. $A = \int_{a}^{b} (top - bottom) d x$
2. $A = \int_{- 2}^{2} (4 - x^{2} - (0)) d x = \frac{32}{3}$
4.
Find the area between the curves.
$y = x + 2, y = x^{2}, - 1 \leq x \leq 2$
Answer: $\frac{9}{2}$
Step-by-step solution
1. $A = \int_{a}^{b} (top - bottom) d x$
2. $A = \int_{- 1}^{2} (x + 2 - (x^{2})) d x = \frac{9}{2}$
5.
Find the area between the curves.
$y = 2 - x, y = x^{2} - 2, - 1 \leq x \leq 1$
Answer: $\frac{22}{3}$
Step-by-step solution
1. $A = \int_{a}^{b} (top - bottom) d x$
2. $A = \int_{- 1}^{1} (2 - x - (x^{2} - 2)) d x = \frac{22}{3}$
6.
Find the area between the curves.
$y = - x^{2} + 3 x, y = x, 0 \leq x \leq 2$
Answer: $\frac{4}{3}$
Step-by-step solution
1. $A = \int_{a}^{b} (top - bottom) d x$
2. $A = \int_{0}^{2} (- x^{2} + 3 x - (x)) d x = \frac{4}{3}$
7.
Find the area between the curves.
$y = 3 x, y = x^{2}, 0 \leq x \leq 3$
Answer: $\frac{9}{2}$
Step-by-step solution
1. $A = \int_{a}^{b} (top - bottom) d x$
2. $A = \int_{0}^{3} (3 x - (x^{2})) d x = \frac{9}{2}$
8.
Find the area between the curves.
$y = x, y = x^{3}, 0 \leq x \leq 1$
Answer: $\frac{1}{4}$
Step-by-step solution
1. $A = \int_{a}^{b} (top - bottom) d x$
2. $A = \int_{0}^{1} (x - (x^{3})) d x = \frac{1}{4}$
9.
Find the area between the curves.
$y = 9 - x^{2}, y = 0, - 3 \leq x \leq 3$
Answer: $36$
Step-by-step solution
1. $A = \int_{a}^{b} (top - bottom) d x$
2. $A = \int_{- 3}^{3} (9 - x^{2} - (0)) d x = 36$
10.
Find the area between the curves.
$y = 4 x, y = x^{2}, 0 \leq x \leq 4$
Answer: $\frac{32}{3}$
Step-by-step solution
1. $A = \int_{a}^{b} (top - bottom) d x$
2. $A = \int_{0}^{4} (4 x - (x^{2})) d x = \frac{32}{3}$
11.
Find the area between the curves.
$y = x + 6, y = x^{2}, - 2 \leq x \leq 3$
Answer: $\frac{125}{6}$
Step-by-step solution
1. $A = \int_{a}^{b} (top - bottom) d x$
2. $A = \int_{- 2}^{3} (x + 6 - (x^{2})) d x = \frac{125}{6}$
12.
Find the area between the curves.
$y = 2 x, y = x^{3}, 0 \leq x \leq 1$
Answer: $\frac{3}{4}$
Step-by-step solution
1. $A = \int_{a}^{b} (top - bottom) d x$
2. $A = \int_{0}^{1} (2 x - (x^{3})) d x = \frac{3}{4}$
13.
Find the area between the curves.
$y = 5 - x^{2}, y = 1, - 2 \leq x \leq 2$
Answer: $\frac{32}{3}$
Step-by-step solution
1. $A = \int_{a}^{b} (top - bottom) d x$
2. $A = \int_{- 2}^{2} (5 - x^{2} - (1)) d x = \frac{32}{3}$
14.
Find the area between the curves.
$y = - x^{2} + 6 x, y = x, 0 \leq x \leq 5$
Answer: $\frac{125}{6}$
Step-by-step solution
1. $A = \int_{a}^{b} (top - bottom) d x$
2. $A = \int_{0}^{5} (- x^{2} + 6 x - (x)) d x = \frac{125}{6}$
15.
Find the area between the curves.
$y = x + 1, y = x^{2} - 1, - 1 \leq x \leq 2$
Answer: $\frac{9}{2}$
Step-by-step solution
1. $A = \int_{a}^{b} (top - bottom) d x$
2. $A = \int_{- 1}^{2} (x + 1 - (x^{2} - 1)) d x = \frac{9}{2}$
16.
Find the area between the curves.
$y = 8 - x^{2}, y = 4, - 2 \leq x \leq 2$
Answer: $\frac{32}{3}$
Step-by-step solution
1. $A = \int_{a}^{b} (top - bottom) d x$
2. $A = \int_{- 2}^{2} (8 - x^{2} - (4)) d x = \frac{32}{3}$

The formulas you need here

Area between two curves

16 worked problems

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