# Integration: Area Under a Curve – A Level Maths Revision

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### Area Under a Curve: Integration

There are two ways of thinking about integration.

Firstly, as the inverse of differentiation (integration in this context is covered in this revision blog).

Secondly, as a way of summing the area’s increasingly small slices under a function in order to find the total area (you can compare this to how differentiation from first principles allows you to find the gradient of a function).

This means that you can use definite integration to find the area under a curve, between two limits (if you’re not sure about definite integration, check out our revision blog on that topic before moving onto this one).

Definite integration allows you to find the area bounded by a curve and the π₯-axis. If the area is above the π₯-axis the area will be positive; if the area is below the π₯-axis the area will be negative.

In the image above, the shaded area can be found by integrating π¦ = f(π₯) with respect to π₯ with an upper limit of π and a lower limit of π:

$\text{Area = }\int^b_a{f(x)}\,dx$

### Example Question 1

Find the area of the region that is bounded between the curve with equation π¦ = 24 – 2π₯2 – 2π₯ and the π₯-axis.

Begin by sketching the curve with equation π¦ = 24 – 2π₯2 – 2π₯ so you can visualise the area you are trying to find.

The curve meets the π₯-axis when π¦ = 0:

\begin{aligned}&0=-2x^2-2x+24\\&0=-2(x+4)(x-3)\\&x=-4\text{ and }x=3 \end{aligned}

This means that, between π₯ = -4 and π₯ = 3, the region is bounded by the curve and the π₯-axis.

You can now find the area under the curve:

\begin{aligned}\text{Area }&=\int^3_{-4}{(24-2x^2-2x)}\,dx \\&= \left[24x-\frac{2}{3}x^3-x^2\right]^3_{-4}\\&=(24(3)-\frac{2}{3}(3)^3-(3)^2)-(24(-4)-\frac{2}{3}(-4)^3-(-4)^2)\\&=\frac{343}{3}\end{aligned}

You have not been given any units, so you do not add any units to your answer.

This answer can be quickly checked or worked out on your calculator using the integration button:

You will still need to make sure that you show full working in exams for method marks.

### Example Question 2

Find the total area of the region bounded by the curve with equation π¦ = π₯3 β 16π₯2 + 63π₯, the π₯-axis, and the lines with equations π₯ = 2 and π₯ = 9.

Again, start by sketching the function.

The curve meets the π₯-axis when π¦ = 0:

\begin{aligned}&0=x^3-16x^2+63x\\&0=x(x-7)(x-9)\\&x=0, x=7\text{ and } x=9 \end{aligned}

You can see that the region is bounded by the curve between the lines with equations π₯ = 2 and π₯ = 9, but that the area between the lines with equations π₯ = 7 and π₯ = 9 will be negative. You want to find the total area so we will need to add the magnitudes of the two areas. (area A and area B on the diagram).

\begin{aligned} \text{Area A }&=\int^7_2{(x^3-16x^2+63x)}\,dx \\ &= \left[\frac{x^4}{4}-\frac{16x^3}{3}+\frac{63x^2}{2}\right]^7_2 \\ &= \left(\frac{1}{4}(7)^4-\frac{16}{3}(7)^3+\frac{63}{2}(7)^2\right)-\left(\frac{1}{4}(2)^4-\frac{16}{3}(2)^3+\frac{63}{2}(2)^2\right)\\&=\frac{2725}{12} \end{aligned}

\begin{aligned} \text{Area B }&=\int^9_7{(x^3-16x^2+63x)}\,dx \\ &= \left[\frac{x^4}{4}-\frac{16x^3}{3}+\frac{63x^2}{2}\right]^9_7 \\ &= \left(\frac{1}{4}(9)^4-\frac{16}{3}(9)^3+\frac{63}{2}(9)^2\right)-\left(\frac{1}{4}(7)^4-\frac{16}{3}(7)^3+\frac{63}{2}(7)^2\right)\\&=-\frac{32}{3} \end{aligned}

The second integration gives a negative answer, but as you are considering area take the absolute (positive) value.

The total area bounded by the curve will be:

\begin{aligned}\text{Area }&=\frac{2725}{12}+\frac{32}{3} \\ &=\frac{951}{4} \end{aligned}

### Example Question 3

Find the area of the finite region bounded between the curve with equation f(π₯) = π₯3 β 20π₯2 + 100π₯ β 225 and the straight-line with equation g(π₯) = -12π₯ β 33.

Again, start by sketching the curve and the line. You are interested in where the functions intercept and whether they cross over the π₯-axis in that range.

Start by finding the intercepts:

\begin{aligned}&x^3-20x^2+100x-225=-12x-33\\&x^3-20x^2+112x-192=0\\&x=4\text{ and }x=12 \\&y=-81\text{ and }y=-177 \end{aligned}

The region you need to calculate the area for is between π₯ = 4 and π₯ = 12.

Values where f(π₯) = 0:
π₯ = 14.0 (3s.f.). This is not within our region.

Values where g(π₯) = 0
π₯ = -2.75. This is not within our region.

You can find the area between the two curves by splitting the area into two separate parts:

This shows $\int^{12}_4{f(x)}\,dx$.

This shows the area above g(π₯). You can calculate this using the formula for the area of a trapezium.

To find the area enclosed between the two curves, subtract the area of the trapezium away from the area between the curve with equation π¦ = f(π₯) and the π₯-axis. Remember, because this region is below the π₯-axis it will have a negative area; use the absolute value of its area to calculate the area of the region between the graphs with equations π¦ = f(π₯) and π¦ = g(π₯).

\begin{aligned} \text{Area above y = f(}x\text{) }&=\int^{12}_4{f(x)}\,dx\\&=\int^{12}_4{(x^3-20x^2+100x-225)}\,dx\\&=\left[\frac{1}{4}x^4-\frac{20}{3}x^3+50x^2-225x\right]^{12}_4\\&=-\frac{4120}{3}\end{aligned}

\begin{aligned} \text{Area above y = g(}x\text{) }&=\frac{1}{2}(81+177)(8)\\&=1032\end{aligned}

\begin{aligned} \text{Area between y = f(}x\text{) and y = g(}x\text{) }&=\frac{4120}{3}-1032\\&=\frac{1024}{3}\end{aligned}

As you have seen in these examples, it is very important to sketch the functions in order to identify where the functions change sign or intersect with other functions. Depending on the question, you may also need to use areas of associated rectangles, triangles, and trapeziums, as well as integrating to find the area required.

### Example Question 4

Find the area between the curve with equation π¦ = 2π₯2 + 3, the π¦-axis and the lines with equations π¦ = 4 and π¦ = 7.

As usual, start by sketching the curve with equation π¦ = 2π₯2 + 3 to highlight the area the question is asking you to find.

You will need to find the coordinates where the curve meets the lines with equations π¦ = 4 and π¦ = 7.

\begin{aligned}&2x^2+3=4\\&2x^2=1\\&x^2=\frac{1}{2}\\&x=\pm\frac{1}{\sqrt{2}}\end{aligned}

\begin{aligned}&2x^2+3=7\\&2x^2=4\\&x^2=2\\&x=\pm\sqrt{2}\end{aligned}

As you are looking at the region bounded by the curve, the π¦-axis and the lines with equations π¦ = 4 and π¦ = 7, we need the positive values of these π₯-coordinates:

$x=\frac{1}{\sqrt{2}}$

$x=\sqrt{2}$

To find the area of the shaded region, treat it like a compound shape. Here is one way to break down the area usefully:

You want to find Area A.

Area A + Area B is a rectangle, 3 by $\sqrt{2}$.

Area B + Area C is the area under the curve with equation y = 2x2 + 3 between the lines π₯ = $\frac{1}{\sqrt{2}}$ and π₯ =$\sqrt{2}$.

Area C is a rectangle, 4 by ($\sqrt{2}-\frac{1}{\sqrt{2}}$).

Use these facts to find the area of each region:

\begin{aligned}\text{Area B + Area C}&=\int^{\sqrt{2}}_{\frac{1}{\sqrt{2}}}{(2x^2+3)}\,dx\\&=\left[\frac{2}{3} x^3+3x\right]^{\sqrt{2}}_{\frac{1}{\sqrt{2}}}\\&=3.771236\text{...}\end{aligned}

\begin{aligned} \text{Area C} &= 4 \left(\sqrt{2}-\frac{1}{\sqrt{2}}\right)\\&=2\sqrt{2}\end{aligned}

\begin{aligned}\text{Area B} &= (\text{Area B}+\text{Area C})-\text{Area C} \\ &=3.771236... - 2\sqrt{2} \\ &= 0.942809 \end{aligned}

\begin{aligned}\text{Area A} &+ \text{Area B} =3\sqrt{2}\\\text{Area A} &= (\text{Area A} + \text{Area B})-\text{Area B}\\&=3\sqrt{2}-0.942809\\&=3.30\text{(3s.f.)}\end{aligned}

Problems like this can be more difficult if the region of interest lies on both sides of the π₯-axis. In situations like this, it can simplify the problem to shift both functions up or down by the same amount β this will not affect the size of the area of interest.

### Practice Questions

For each question, sketch the curve then find the area specified.

1) The area under the curve with equation $y = \frac{5}{x^2}$ between the lines with equations π₯ = 2 and π₯ = 5.

2) The area bounded by the curve with equation π¦ = (3 β 2π₯)(π₯ + 2) and the π₯-axis.

3) The total area enclosed between the curve with the equation π¦ = 6π₯3 β 11π₯2 β 35π₯ and the π₯-axis.

4) The total area enclosed between the curve with equation f(π₯) = 2π₯3 β 33π₯2 + 133π₯, the π₯-axis and the lines with equations π₯ = 2 and π₯ = 12.

5) The area between the curves with equations f(π₯) = π₯2 + 4π₯ + 5 and g(π₯) = $\frac{1}{2}x$ + 7.

6) The area of the region bounded by the curve with equation π¦ = π₯2 + π₯ + 1 and the lines with equations π¦ = 3 and π¦ = 7.

1) The area under the curve with equation $y = \frac{5}{x^2}$ between π₯ = 2 and π₯ =5.

\boldsymbol{\begin{aligned} \textbf{Area} &= \int^5_2{(5x^{-2})}\,dx\\&=\left[-5x^{-1}\right]^5_2\\&=1.5\end{aligned}}

2) The area bounded by the curve with equation π¦ = (3 β 2π₯)(π₯ + 2) and the π₯-axis.

$\boldsymbol{y=-2x^2-x+6 }$

\boldsymbol{\begin{aligned} \textbf{Area} &= \int^{1.5}_{-2}{(-2x^2-x+6)}\,dx\\ &= \left[ -\frac{2}{3}x^3-\frac{1}{2}x^2+6x\right]^{1.5}_{-2} \\ &=\frac{343}{24} \end{aligned}}

3) The total area enclosed between the curve with equation π¦ = 6π₯3 β 11π₯2 β 35π₯ and the π₯-axis.

\boldsymbol{\begin{aligned}&y=x(3x+5)(2x-7)\\&\textbf{Crosses at } x = -\frac{5}{3}, x = 0 \textbf { and } x = \frac{7}{2} \end{aligned}}

\boldsymbol{\begin{aligned}\textbf{Area A}&=\int^0_{-\frac{5}{3}}{(6x^3-11x^2-35x)}\,dx\\&=\left[\frac{3}{2}x^4-\frac{11}{3}x^3-\frac{35}{2}x^2\right]^0_{-\frac{5}{3}}\\&=\frac{1625}{81}\end{aligned}}

\boldsymbol{\begin{aligned}\textbf{Area B}&=\int^{\frac{7}{2}}_0{(6x^3-11x^2-35x)}\,dx\\&=\left[\frac{3}{2}x^4-\frac{11}{3}x^3-\frac{35}{2}x^2\right]^{\frac{7}{2}}_{0}\\ &=-\frac{14 063}{96}\end{aligned}}

\boldsymbol{\begin{aligned}\textbf{Total Area} &= \frac{1625}{81}+\frac{14 063}{96} \\ &= 166.5513 \\ &= 167 \textbf{ (3s.f.)}\end{aligned}}

4) The total area enclosed between the curve with equation f(π₯) = 2π₯3 β 33π₯2 + 133π₯, the π₯-axis and the lines with equations π₯ = 2 and π₯ = 12.

\boldsymbol{\begin{aligned}&f(x)=x(x-7)(2x-19)\\&=\textbf{Crosses at }x=0,x=7 \textbf{ and } x=\frac{19}{2}\end{aligned}}

\boldsymbol{\begin{aligned}\textbf{Area A}&=\int^7_2{(2x^3-33x^2+133x)}\, dx \\ &=\left[\frac{1}{2}x^4-11x^3+\frac{133}{2}x^2\right]^7_2\\&=500\end{aligned}}

\boldsymbol{\begin{aligned}\textbf{Area B}&=\int^{9.5}_7{(2x^3-33x^2+133x)}\, dx\\&=\left[\frac{1}{2}x^4-11x^3+\frac{133}{2}x^2\right]^{9.5}_{7}\\&=-\frac{1375}{32}\end{aligned}}

\boldsymbol{\begin{aligned}\textbf{Area C}&=\int^{12}_{9.5}{(2x^3-33x^2+133x)}\, dx\\&=\left[\frac{1}{2}x^4-11x^3+\frac{133}{2}x^2\right]^{12}_{9.5}\\&=\frac{9375}{32}\end{aligned}}

\boldsymbol{\begin{aligned}\textbf{Total Area}&=500+\frac{1375}{32}+\frac{9375}{32}\\&=\frac{13375}{16}\\&=836\textbf{ (3s.f.)} \end{aligned}}

5) The area between the curves with equations f(π₯) = π₯2 + 4π₯ + 5 and g(π₯) = $\frac{1}{2}x$ + 7.

\boldsymbol{\begin{aligned}&\textbf{Find points of intersection:}\\&x^2+4x+5=\frac{1}{2}x+7\\&x^2+3.5x-2=0\\&x=-4\textbf{ and }x=0.5\\&(-4,5)\textbf{ and }(\frac{1}{2}, \frac{29}{4}) \end{aligned}}

\boldsymbol{\begin{aligned}\textbf{Area} &\textbf{ of trapezium below }g(x)\textbf{:}\\ \textbf{Area} &=\frac{1}{2}\left(5+\frac{29}{4}\right)\left(4.5\right)\\&=\frac{441}{16}\end{aligned}}

\boldsymbol{\begin{aligned}\textbf{Area} &\textbf{ below }f(x)\textbf{:}\\\textbf{Area}&=\int^{0.5}_{-4}{(x^2+4x+5)}\,dx\\&=\left[\frac{1}{3}x^3+2x^2+5x\right]^{0.5}_{-4}\\&=\frac{99}{8}\end{aligned}}

\boldsymbol{\begin{aligned}&\textbf{Area between }f(x) \textbf{ and }g(x)\textbf{:}\\&\frac{441}{16}-\frac{99}{8}=\frac{243}{16}\end{aligned}}

6) The area of the region bounded by the curve with equation π¦ = π₯2 + π₯ + 1 and the lines with equations π¦ = 3 and π¦ = 7.

\boldsymbol{\begin{aligned}&\textbf{Points of intersection:}\\&x^2+x+1=3\\&x^2+x-2=0\\&x=-2\textbf{ or }x=1\\&=(-2,3)\textbf{ and }(1,3)\end{aligned}}

\boldsymbol{\begin{aligned}&x^2+x+1=7\\&x^2+x-6=0\\&x=-3\textbf{ and }x=2\\&(-3,7)\textbf{ and }(2,7)\end{aligned}}

\boldsymbol{\begin{aligned}\textbf{Area A + B}&=\int^{-2}_{-3}{y}\,dx\\&=\int^{-2}_{-3}{(x^2+x+1)}\,dx\\&=\frac{29}{6}\end{aligned}}

\boldsymbol{\begin{aligned}\textbf{Area D + E}&=\int^{2}_{1}{y}\,dx\\&=\int^{2}_{1}{(x^2+x+1)}\,dx\\&=\frac{29}{6}\end{aligned}}

\boldsymbol{\begin{aligned}&\textbf{Area A, Area E and (Area B + C + D) are rectangles.}\\&\textbf{Area A} = 3\\&\textbf{Area E} = 3\\&\textbf{Area B + C + D} = 20\end{aligned}}

\boldsymbol{\begin{aligned}\textbf{Area B}&= \textbf{(Area A + B) - Area A}\\&=\frac{11}{6}\end{aligned}}

\boldsymbol{\begin{aligned}\textbf{Area D}&= \textbf{(Area D + E) - Area E}\\&=\frac{11}{6}\end{aligned}}

\boldsymbol{\begin{aligned}\textbf{Area C}&= \textbf{(Area B + C + D) - Area B - Area D}\\&=20-\frac{11}{6}-\frac{11}{6}\\&=\frac{49}{3}\end{aligned}}

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