Finding the Area Between Curves Expressed as Functions of x

AP Calculus · Applications of Integration

Area Between Curves as Functions of x

One vertical slice, height top minus bottom, width dx. The integral does the rest.

key idea$A=\int_a^b\Big[\text{top}(x)-\text{bottom}(x)\Big]dx$

Picture the region

Draw the region and drop one vertical slice through it. That slice runs from the lower curve up to the upper curve, so its height is $\text{top}-\text{bottom}$ and its width is $dx$. Summing those thin rectangles is exactly what the integral does.

The region between $y=4-x^{2}$ and $y=x+2$. The dots mark where the curves meet, at $x=-2$ and $x=1$ — those are the limits of integration, and they come from solving the two equations against each other, not from the picture's edges.

Setting up the integral

From a picture to a definite integral

Sketch both curves and shade the region. Two minutes here saves ten later.
Find the limits by setting the curves equal and solving. Only use given endpoints when the problem supplies them explicitly.
Decide which curve is on top by testing a single convenient $x$ strictly inside the interval — substitute it into both.
Write the height as $\text{top}-\text{bottom}$, in that order. Getting it backwards gives the right magnitude with the wrong sign.
Integrate over $[a,b]$ and check that the answer is positive. An area that comes out negative means the two curves were swapped.

Situation	Height of the slice	Note
both curves above the axis	$f-g$	positions relative to the axis are irrelevant
both below the axis	$f-g$	the same formula — the subtraction handles it
one above, one below	$f-g$	still the same; $f-g$ is the vertical gap
region bounded by the $x$-axis	$f-0=f$	the axis is the line $y=0$
curves cross inside $[a,b]$	—	split the integral at the crossing

Only the last row changes the method. The first four are the same integral: the vertical gap between the curves does not care where the $x$-axis is.

the four decisions

The limits come from solving $f=g$, not from reading the picture's edges.
Test one interior point to decide which curve is on top — guessing from a rough sketch is unreliable.
$\text{top}-\text{bottom}$ works whether the curves are above, below, or straddling the axis.
If the curves cross inside the interval, split there and add the pieces.

Worked examples

EX 1

Parabola and line

the standard set-up

Find the area between $y=4-x^{2}$ and $y=x+2$.

Find the intersections

$4-x^{2}=x+2$ gives $x^{2}+x-2=0$, so $(x+2)(x-1)=0$ and $x=-2,\,1$.

Decide which is on top

At $x=0$: the parabola gives $4$, the line gives $2$. The parabola is on top.

Write the integral

$A=\displaystyle\int_{-2}^{1}\left[\left(4-x^{2}\right)-(x+2)\right]dx=\int_{-2}^{1}\left(2-x-x^{2}\right)dx$.

Antidifferentiate

$\left[2x-\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}\right]_{-2}^{1}$.

Evaluate

$\left(2-\tfrac12-\tfrac13\right)-\left(-4-2+\tfrac83\right)=\dfrac76+\dfrac{10}{3}=\dfrac92$.

EX 2

Region below the axis

the formula does not change

Find the area between $y=x^{2}-4$ and $y=-3$ on the interval where the first lies below the second.

Find the intersections

$x^{2}-4=-3$ gives $x^{2}=1$, so $x=\pm1$.

Test an interior point

At $x=0$: $x^{2}-4=-4$ and $-3$. So $y=-3$ is on top — even though both are negative.

Write the integral

$A=\displaystyle\int_{-1}^{1}\left[-3-\left(x^{2}-4\right)\right]dx=\int_{-1}^{1}\left(1-x^{2}\right)dx$.

Evaluate

$\left[x-\dfrac{x^{3}}{3}\right]_{-1}^{1}=\dfrac23-\left(-\dfrac23\right)=\dfrac43$.

Note the point

Being below the $x$-axis never enters the calculation. Only which curve is higher matters.

EX 3

Bounded by the axis

the axis is a curve

Find the area enclosed by $y=x^{2}-9$ and the $x$-axis.

Treat the axis as $y=0$

The two boundaries are $y=x^{2}-9$ and $y=0$.

Find the limits

$x^{2}-9=0$ gives $x=\pm3$.

Decide which is on top

At $x=0$: the parabola gives $-9$, the axis gives $0$. The axis is on top.

Set up and integrate

$A=\displaystyle\int_{-3}^{3}\left(0-\left(x^{2}-9\right)\right)dx=\int_{-3}^{3}\left(9-x^{2}\right)dx$.

Evaluate

$\left[9x-\dfrac{x^{3}}{3}\right]_{-3}^{3}=18-(-18)=36$. Writing $\int\left(x^{2}-9\right)$ instead would have produced $-36$.

EX 4

A trigonometric region

limits from an equation

Find the area between $y=\sin x$ and $y=\cos x$ on $\left[\dfrac{\pi}{4},\dfrac{5\pi}{4}\right]$.

Confirm the endpoints are intersections

$\sin x=\cos x$ at $x=\dfrac{\pi}{4}$ and $x=\dfrac{5\pi}{4}$, so the region closes at both ends.

Test an interior point

At $x=\dfrac{\pi}{2}$: $\sin=1$, $\cos=0$. Sine is on top throughout.

Write the integral

$A=\displaystyle\int_{\pi/4}^{5\pi/4}\left(\sin x-\cos x\right)dx$.

Antidifferentiate

$\left[-\cos x-\sin x\right]_{\pi/4}^{5\pi/4}$.

Evaluate

$\left(\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}\right)-\left(-\tfrac{\sqrt2}{2}-\tfrac{\sqrt2}{2}\right)=2\sqrt2$.

Watch out

Solve for the limits — do not read them off

The endpoints are where $f=g$. A hand sketch is for orientation; the algebra supplies the numbers.

A negative answer means the order was wrong

Area is positive. If the arithmetic gives a negative number, you subtracted top from bottom — swap and keep the magnitude.

Position relative to the axis is irrelevant

$\text{top}-\text{bottom}$ is the vertical gap. Regions entirely below the axis use the identical formula.

Check whether the curves cross inside the interval

One crossing turns a single integral into two. Test a point in each sub-interval before committing to a set-up.

Course map. Topic 8.4 · LO CHA-5.A — calculate areas in the plane using the definite integral.