Application of Integrals • Topic 2 of 2

Area Between Two Curves

The area enclosed between two curves $y=f(x)$ (the upper curve) and $y=g(x)$ (the lower curve) over $[a,b]$ is the integral of the vertical gap between them:

$$\text{Area}=\int_a^b \big[f(x)-g(x)\big]\,dx,\qquad f(x)\ge g(x)\ \text{on }[a,b].$$

Each vertical strip now runs from the lower curve up to the upper curve, so its height is $(\text{top}-\text{bottom})$ and its area is $\big[f(x)-g(x)\big]\,dx$. The method is always the same three steps.

Sketch both curves and shade the enclosed region.
Find the limits from the points of intersection: solve $f(x)=g(x)$. The $x$-coordinates of the intersections are $a$ and $b$.
Integrate top minus bottom. Test a sample point inside the interval to decide which curve is upper, so the integrand stays non-negative.

If the curves cross inside the interval

When the two curves swap top and bottom partway through, split the interval at every crossing and integrate each sub-interval with its own correct "upper minus lower" order. The total area is the sum of these positive pieces:

$$\text{Area}=\int_a^c\big[f-g\big]\,dx+\int_c^b\big[g-f\big]\,dx.$$

Integrating along the $y$-axis

Sometimes "right curve minus left curve" with horizontal strips is far cleaner — especially for a parabola opening sideways together with a line. Then

$$\text{Area}=\int_c^d \big[x_{\text{right}}(y)-x_{\text{left}}(y)\big]\,dy.$$

Area of a triangle from its sides or vertices

A triangle whose three sides are given as line equations can be found by integration. Find the three vertices (intersections of the lines taken in pairs), then split the base interval at the $x$-coordinate where the upper boundary switches from one line to another, and integrate "upper line minus lower line" over each part. For example, for the triangle with vertices $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$ ordered by $x$, integrate the top edge minus the bottom edge on $[x_1,x_2]$ and on $[x_2,x_3]$. The integral answer should agree with the determinant formula $\dfrac12\,|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$ — a handy check.

Regions involving modulus (absolute-value) graphs

A graph such as $y=|x|$ or $y=|x-1|$ is built from two straight pieces with a corner where the inside changes sign. To find an area involving a modulus, break it at the corner: write $|x-k|=x-k$ for $x\ge k$ and $|x-k|=-(x-k)$ for $x

Deeper Insight — one principle behind every variant: top minus bottom. Whether the boundaries are two parabolas, a circle and a line, or two straight edges of a triangle, the area is always $\int(\text{upper}-\text{lower})$. The only real work is the sketch: it tells you the limits (the intersections), which curve is on top in each stretch, and whether the roles swap (so you must split) or whether horizontal strips are easier. A modulus graph is not a new case — it is just a function that changes its formula at a corner, so you split there exactly as you would at any crossing. Resist the urge to plug numbers before drawing; nearly every error in this topic is a wrong limit or a flipped sign that a quick, honest sketch would have caught.

Solved Examples

Worked Example

Find the area between $y=x$ and $y=x^2$ from their intersections.

Solution

Intersections: $x=x^2\Rightarrow x=0,1$. On $(0,1)$, $x\ge x^2$. Area $=\int_0^1 (x-x^2)\,dx=\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1=\dfrac12-\dfrac13=\dfrac16$ square units.

Answer: $\dfrac16$ square units.

Worked Example

Find the area between $y=x^2$ and $y=4$.

Solution

Intersections: $x^2=4\Rightarrow x=\pm2$. Upper curve is $y=4$. Area $=\int_{-2}^{2}(4-x^2)\,dx=\left[4x-\dfrac{x^3}{3}\right]_{-2}^{2}=\left(8-\dfrac83\right)-\left(-8+\dfrac83\right)=\dfrac{32}{3}$ square units.

Answer: $\dfrac{32}{3}$ square units.

Worked Example

Find the area between the line $y=2x$ and the parabola $y=x^2$.

Solution

$x^2=2x\Rightarrow x=0,2$. On $(0,2)$, $2x\ge x^2$. Area $=\int_0^2(2x-x^2)\,dx=\left[x^2-\dfrac{x^3}{3}\right]_0^2=4-\dfrac83=\dfrac{4}{3}$ square units.

Answer: $\dfrac{4}{3}$ square units.

Worked Example

Set up the area between $y=\sqrt{x}$ and $y=x$ for $0\le x\le1$.

Solution

On $(0,1)$, $\sqrt{x}\ge x$. Area $=\int_0^1(\sqrt{x}-x)\,dx=\left[\dfrac{2}{3}x^{3/2}-\dfrac{x^2}{2}\right]_0^1=\dfrac23-\dfrac12=\dfrac16$ square units.

Answer: $\dfrac16$ square units.

Worked Example

Find the area of the region bounded by the parabola $y^2=4x$ and the line $y=2x$.

Solution

Solve: from $y=2x$, $y^2=4x^2$; set equal to $4x$ to get $4x^2=4x\Rightarrow x=0,1$, giving points $(0,0)$ and $(1,2)$. Use horizontal strips, $0\le y\le 2$: the parabola gives $x=\dfrac{y^2}{4}$ (right boundary... check) — test $y=1$: parabola $x=\tfrac14$, line $x=\tfrac{y}{2}=\tfrac12$, so the line is to the right. Area $=\int_0^2\left(\dfrac{y}{2}-\dfrac{y^2}{4}\right)dy=\left[\dfrac{y^2}{4}-\dfrac{y^3}{12}\right]_0^2=1-\dfrac{8}{12}=1-\dfrac23=\dfrac13$ square unit.

Answer: $\dfrac13$ square unit.

Worked Example

Find the area enclosed between the parabola $y=x^2$ and the line $y=x+2$.

Solution

Intersections: $x^2=x+2\Rightarrow x^2-x-2=0\Rightarrow (x-2)(x+1)=0\Rightarrow x=-1,2$. On $(-1,2)$ the line is above the parabola. Area $=\int_{-1}^{2}\big[(x+2)-x^2\big]\,dx=\left[\dfrac{x^2}{2}+2x-\dfrac{x^3}{3}\right]_{-1}^{2}$. At $x=2$: $2+4-\dfrac83=\dfrac{10}{3}$. At $x=-1$: $\dfrac12-2+\dfrac13=-\dfrac{7}{6}$. Difference $=\dfrac{10}{3}+\dfrac{7}{6}=\dfrac{27}{6}=\dfrac92$ square units.

Answer: $\dfrac92$ square units.

Worked Example

Find the area common to the two parabolas $y^2=x$ and $x^2=y$.

Solution

Intersections: substitute $y=x^2$ into $y^2=x$: $x^4=x\Rightarrow x(x^3-1)=0\Rightarrow x=0,1$, points $(0,0)$ and $(1,1)$. On $(0,1)$ the upper curve is $y=\sqrt{x}$ (from $y^2=x$) and the lower is $y=x^2$. Area $=\int_0^1(\sqrt{x}-x^2)\,dx=\left[\dfrac{2}{3}x^{3/2}-\dfrac{x^3}{3}\right]_0^1=\dfrac23-\dfrac13=\dfrac13$ square unit.

Answer: $\dfrac13$ square unit.

Worked Example

Find the area of the smaller region bounded by the circle $x^2+y^2=4$ and the line $x=1$ (the region to the right of the line).

Solution

The chord $x=1$ cuts the circle at $y=\pm\sqrt{3}$. For the region $1\le x\le 2$, the curve gives $y=\pm\sqrt{4-x^2}$, so a vertical strip has height $2\sqrt{4-x^2}$. Area $=\int_1^2 2\sqrt{4-x^2}\,dx=2\left[\dfrac{x}{2}\sqrt{4-x^2}+\dfrac{4}{2}\sin^{-1}\dfrac{x}{2}\right]_1^2$. At $x=2$: $2\left(0+2\cdot\dfrac{\pi}{2}\right)=2\pi$. At $x=1$: $2\left(\dfrac12\sqrt{3}+2\cdot\dfrac{\pi}{6}\right)=\sqrt{3}+\dfrac{2\pi}{3}$. Area $=2\pi-\sqrt{3}-\dfrac{2\pi}{3}=\dfrac{4\pi}{3}-\sqrt{3}$ square units.

Answer: $\dfrac{4\pi}{3}-\sqrt{3}$ square units.

Worked Example

Find the area of the region bounded by the parabola $y^2=4x$ and its latus rectum (the line $x=1$).

Solution

For $y^2=4x$, $4a=4$ so $a=1$ and the latus rectum is $x=1$, meeting the curve at $(1,\pm2)$. A vertical strip on $0\le x\le 1$ runs from $y=-2\sqrt{x}$ to $y=2\sqrt{x}$, height $4\sqrt{x}$. Area $=\int_0^1 4\sqrt{x}\,dx=4\left[\dfrac{2}{3}x^{3/2}\right]_0^1=\dfrac{8}{3}$ square units.

Answer: $\dfrac{8}{3}$ square units.

Worked Example

Using integration, find the area of the triangle with vertices $A(1,0)$, $B(2,2)$ and $C(3,1)$.

Solution

Sides as lines: $AB:y=2(x-1)=2x-2$; $AC:y=\dfrac{x-1}{2}$; $BC:y=-x+4$. Split at $x=2$. On $[1,2]$ the top is $AB$ and bottom is $AC$; on $[2,3]$ the top is $BC$ and bottom is $AC$.
$\int_1^2\!\left[(2x-2)-\dfrac{x-1}{2}\right]dx=\int_1^2\dfrac{3(x-1)}{2}\,dx=\dfrac{3}{2}\left[\dfrac{(x-1)^2}{2}\right]_1^2=\dfrac34$.
$\int_2^3\!\left[(-x+4)-\dfrac{x-1}{2}\right]dx=\int_2^3\dfrac{9-3x}{2}\,dx=\dfrac{1}{2}\left[9x-\dfrac{3x^2}{2}\right]_2^3=\dfrac{1}{2}\left(\dfrac{27}{2}-12\right)=\dfrac34$.
Total area $=\dfrac34+\dfrac34=\dfrac32$ square units. (Check by determinant: $\dfrac12|1(2-1)+2(1-0)+3(0-2)|=\dfrac12|1+2-6|=\dfrac32$.)

Answer: $\dfrac32$ square units.

Worked Example

Find the area of the region bounded by $y=|x|$ and $y=2$ (the region below the line and above the V).

Solution

The corner of $y=|x|$ is at the origin; it meets $y=2$ at $x=\pm2$. By symmetry about the $y$-axis, area $=2\int_0^2(2-x)\,dx=2\left[2x-\dfrac{x^2}{2}\right]_0^2=2(4-2)=4$ square units.

Answer: $4$ square units.

Worked Example

Find the area enclosed by $|x|+|y|=2$.

Solution

The graph is a square with vertices $(\pm2,0)$ and $(0,\pm2)$. In the first quadrant the edge is $x+y=2$, i.e. $y=2-x$, $0\le x\le 2$. By four-fold symmetry, area $=4\int_0^2(2-x)\,dx=4\left[2x-\dfrac{x^2}{2}\right]_0^2=4(4-2)=8$ square units. (Check: a square of diagonal $2a$ with $a=2$ has area $2a^2=8$.)

Answer: $8$ square units.

Key Points

Area between two curves $=\int_a^b(\text{upper}-\text{lower})\,dx$, summing vertical strips of height $(\text{top}-\text{bottom})$.
The limits $a,b$ are the $x$-coordinates of the intersections: solve $f(x)=g(x)$ first.
Test a sample point inside the interval to decide which curve is upper, so the integrand stays non-negative.
If the curves cross inside the interval, split there and add the positive pieces — never integrate straight through a swap.
Horizontal strips $\int_c^d(x_{\text{right}}-x_{\text{left}})\,dy$ are often cleaner for a sideways parabola with a line.
A triangle from its side-lines: find the vertices, split at the middle $x$, integrate top edge minus bottom edge; cross-check with the determinant area.
Modulus graphs change formula at a corner — break $|x-k|$ into its two branches and integrate each, then add.
For $|x|+|y|=a$ the region is a square of area $2a^2$; use the four-fold symmetry to integrate one piece.

Quick Check — 4 MCQs

Tap an option to check your answer0 / 4

Q1.Area between $y=x$ and $y=x^2$ on $[0,1]$ is:

Explanation: $\int_0^1(x-x^2)dx=1/6$.

Q2.The limits for area between two curves come from:

Explanation: Solve $f(x)=g(x)$.

Q3.Area between $y=x^2$ and $y=4$ is:

Explanation: $\int_{-2}^{2}(4-x^2)dx=32/3$.

Q4.The integrand for area between curves should be:

Explanation: Upper minus lower keeps it non-negative.

Practice more MCQs → 📝 Assignment · 35 marks