Sec 4 A-Math · Definite integrals and areas

Areas questions with answers

Practice questions on areas, from the basics up to exam level, each with a full worked solution. Try them on paper first, then open the solution to check your working.

Worked example

Find the area of the region bounded by the curve $y = -x^{2} + 8x - 7$ , the $x$ -axis and the lines $x = 3$ and $x = 6$ .

Step 1 of 1

Integrate y between the limits, then upper limit minus lower limit

\int_{3}^{6} \left(-x^{2} + 8x - 7\right) \, dx = \Big[-\dfrac{1}{3}x^{3} + 4x^{2} - 7x\Big]_{3}^{6}

= \Big(30\Big) - \Big(6\Big)

= 24

units²

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Questions

Question 1Above the x-axis
Find the area of the region bounded by the curve $y = -x^{2} + 9$ , the $x$ -axis and the lines $x = -2$ and $x = 0$ .
Show the worked solution
1. Step 1: Integrate y between the limits, then upper limit minus lower limit $\int_{-2}^{0} \left(-x^{2} + 9\right) \, dx = \Big[-\dfrac{1}{3}x^{3} + 9x\Big]_{-2}^{0}$ $= \Big(0\Big) - \Big(-\dfrac{46}{3}\Big)$ $= \dfrac{46}{3}$ units²
Question 2Below the x-axis
Find the area of the region bounded by the curve $y = x^{2} - 4x - 5$ , the $x$ -axis and the lines $x = -1$ and $x = 1$ .
Show the worked solution
1. Step 1: Integrate y between the limits, then upper limit minus lower limit $\int_{-1}^{1} \left(x^{2} - 4x - 5\right) \, dx = \Big[\dfrac{1}{3}x^{3} - 2x^{2} - 5x\Big]_{-1}^{1}$ $= \Big(-\dfrac{20}{3}\Big) - \Big(\dfrac{8}{3}\Big)$ $= -\dfrac{28}{3}$
2. Step 2: Is the region above or below the x-axis? What does that mean for the area?The region is below the x-axis, so the integral is negative: the area is its size.
3. Step 3: So the area is $\text{Area} = \dfrac{28}{3}$ units²
Question 3Enclosed by a curve and the x-axis
Find the area of the region enclosed by the curve $y = x^{2} - 4x$ and the $x$ -axis.
Show the worked solution
1. Step 1: Find where the curve meets the x-axis: set y = 0 and factorise $x(x - 4) = 0$ $x = 0$ or $x = 4$
2. Step 2: Integrate y between the two points where it meets the x-axis (they are the limits), then upper limit minus lower limit $\int_{0}^{4} \left(x^{2} - 4x\right) \, dx = \Big[\dfrac{1}{3}x^{3} - 2x^{2}\Big]_{0}^{4}$ $= \Big(-\dfrac{32}{3}\Big) - \Big(0\Big)$ $= -\dfrac{32}{3}$
3. Step 3: Is the region above or below the x-axis? What does that mean for the area?The region is below the x-axis, so the integral is negative: the area is its size.
4. Step 4: So the area is $\text{Area} = \dfrac{32}{3}$ units²
Question 4Between a curve and a line
Find the area of the region enclosed by the curve $y = x^{2} - 6x + 3$ and the line $y = 3$ .
Show the worked solution
1. Step 1: Where do they meet? Set the curve equal to the line, bring everything to one side and factorise $x^{2} - 6x + 3 = 3$ $x^{2} - 6x = 0$ $x(x - 6) = 0$ $x = 0$ or $x = 6$
2. Step 2: Which is on top between the meeting points?Between the meeting points the line is above the curve.
3. Step 3: Area = ∫ (top − bottom) dx, with the meeting points as the limits $\text{Area} = \int_{0}^{6} \left(-x^{2} + 6x\right) \, dx$
4. Step 4: Integrate, then upper limit minus lower limit $= \Big[-\dfrac{1}{3}x^{3} + 3x^{2}\Big]_{0}^{6}$ $= \Big(36\Big) - \Big(0\Big)$ $= 36$ units²

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Worked example

Find the area of the region bounded by the curve

y = -x^{2} + 8x - 7

, the

x

-axis and the lines

x = 3

and

x = 6

Step 1 of 1

Integrate y between the limits, then upper limit minus lower limit

\int_{3}^{6} \left(-x^{2} + 8x - 7\right) \, dx = \Big[-\dfrac{1}{3}x^{3} + 4x^{2} - 7x\Big]_{3}^{6}

= \Big(30\Big) - \Big(6\Big)

= 24

units²

Questions

Question 1Above the x-axis

Find the area of the region bounded by the curve $y = -x^{2} + 9$ , the $x$ -axis and the lines $x = -2$ and $x = 0$ .

Show the worked solution

Step 1: Integrate y between the limits, then upper limit minus lower limit $\int_{-2}^{0} \left(-x^{2} + 9\right) \, dx = \Big[-\dfrac{1}{3}x^{3} + 9x\Big]_{-2}^{0}$ $= \Big(0\Big) - \Big(-\dfrac{46}{3}\Big)$ $= \dfrac{46}{3}$ units²

Question 2Below the x-axis

Find the area of the region bounded by the curve $y = x^{2} - 4x - 5$ , the $x$ -axis and the lines $x = -1$ and $x = 1$ .

Show the worked solution

Step 1: Integrate y between the limits, then upper limit minus lower limit $\int_{-1}^{1} \left(x^{2} - 4x - 5\right) \, dx = \Big[\dfrac{1}{3}x^{3} - 2x^{2} - 5x\Big]_{-1}^{1}$ $= \Big(-\dfrac{20}{3}\Big) - \Big(\dfrac{8}{3}\Big)$ $= -\dfrac{28}{3}$
Step 2: Is the region above or below the x-axis? What does that mean for the area?The region is below the x-axis, so the integral is negative: the area is its size.
Step 3: So the area is $\text{Area} = \dfrac{28}{3}$ units²

Question 3Enclosed by a curve and the x-axis

Find the area of the region enclosed by the curve $y = x^{2} - 4x$ and the $x$ -axis.

Show the worked solution

Step 1: Find where the curve meets the x-axis: set y = 0 and factorise $x(x - 4) = 0$ $x = 0$ or $x = 4$
Step 2: Integrate y between the two points where it meets the x-axis (they are the limits), then upper limit minus lower limit $\int_{0}^{4} \left(x^{2} - 4x\right) \, dx = \Big[\dfrac{1}{3}x^{3} - 2x^{2}\Big]_{0}^{4}$ $= \Big(-\dfrac{32}{3}\Big) - \Big(0\Big)$ $= -\dfrac{32}{3}$
Step 3: Is the region above or below the x-axis? What does that mean for the area?The region is below the x-axis, so the integral is negative: the area is its size.
Step 4: So the area is $\text{Area} = \dfrac{32}{3}$ units²

Question 4Between a curve and a line

Find the area of the region enclosed by the curve $y = x^{2} - 6x + 3$ and the line $y = 3$ .

Show the worked solution

Step 1: Where do they meet? Set the curve equal to the line, bring everything to one side and factorise $x^{2} - 6x + 3 = 3$ $x^{2} - 6x = 0$ $x(x - 6) = 0$ $x = 0$ or $x = 6$
Step 2: Which is on top between the meeting points?Between the meeting points the line is above the curve.
Step 3: Area = ∫ (top − bottom) dx, with the meeting points as the limits $\text{Area} = \int_{0}^{6} \left(-x^{2} + 6x\right) \, dx$
Step 4: Integrate, then upper limit minus lower limit $= \Big[-\dfrac{1}{3}x^{3} + 3x^{2}\Big]_{0}^{6}$ $= \Big(36\Big) - \Big(0\Big)$ $= 36$ units²