1.5 Area between curves

When we say “area between," we want positive area, not signed area.

Area between curves $\approx \frac{\pi}{4}\left(2+\sqrt{2}\right)$

The intervals of our rectangles are $[0,\frac{\pi}{4}]$, $[\frac{\pi}{4},\frac{\pi}{2}]$, $[\frac{\pi}{2},\frac{3\pi}{4}]$, and $[\frac{3\pi}{4},\pi]$. Since we're taking a left Riemann sum, we find the height of the rectangles at the left endpoints of the intervals.

So, our approximation for the area between the two curves is

We're taking rectangles that reach from one function to the other.

(a) Vertical rectangles:

(b) One possible answer:

Draw a sketch first.

$\displaystyle\int_0^{\sqrt{2}}\big[2x-x^3\big]\ \dee{x}$

The curves intersect when $y=x$ and $y=x^3-x$. To find these points, we set:

For $x\ge 0$, the curves intersect at $(0,0)$ and $(\sqrt{2},\sqrt{2})$.

A handy observation is that, since both curves are continuous and they do not meet each other between $x=0$ and $x=\sqrt{2}$, we don't have to worry about dividing our area into two regions: one of the functions is always on the top, and the other is always on the bottom.

Using vertical strips:

The top and bottom boundaries of the specified region are $y=T(x)=x$ and $y=B(x)=x^3-x$, respectively. So,

Draw a sketch first.

$\displaystyle \int_{-3/2}^{4}\left[\frac{4}{5}(6-y^2)+2y\right]\ \dee{y}$

We need to find where the curves intersect.

The curves intersect at $(-8,4)$ and $(3,-\frac{3}{2})$. Using horizontal strips:

we have

You can probably find the intersections by inspection.

$\displaystyle\int_0^{4a}\left[\sqrt{4ax}-\frac{x^2}{4a}\right]\ \dee{x}$

If the curves intersect at $(x,y)$, then

The curves intersect at $(0,0)$ and $(4a,4a)$. (It is also possible to find these points by inspection.) Using vertical strips:

We want the $y$-values of the functions. We write the top function as $y =\sqrt{4ax}$ (we care about the positive square root, not the negative one) and we write the bottom function as $y=\frac{x^2}{4a}$. Then we have

To find the intersection, plug $x=4y^2$ into the equation $x+12y+5=0$.

$\displaystyle\int_1^{25}\left[-\frac{1}{12}(x+5)+\frac{1}{2}\sqrt{x}\right]\ \dee{x}$

The curves intersect when $x=4y^2$ and $0=4y^2+12y+5 =(2y+5)(2y+1)$. So, the curves intersect at $(1,-\half)$ and $(25,-\frac{5}{2})$. Using vertical strips:

we have

If the bottom function is the $x$-axis, this is a familiar question.

$\dfrac{1}{8}$

The area between the curve $y= \frac{1}{(2x-4)^2}$ and the $x$-axis, with $x$ running from $a=0$ to $b=1$, is exactly the definite integral of $\frac{1}{(2x-4)^2}$ with limits $0$ and $1$.

Part of the job is to determine whether $y=x$ lies above or below $y=3x-x^2$.

$\dfrac{4}{3}$

If the curves $y=f(x)=x$ and $y=g(x)=3x-x^2$ intersect at $(x,y)$, then

Furthermore, $g(x)-f(x) = 2x-x^2 = x(2-x)$ is positive for all $0\le x\le 2$. That is, the curve $y=3x-x^2$ lies above the line $y=x$ for all $0\le x\le 2$.

We therefore evaluate the integral:

Guess the intersection points by trying small integers.

$\dfrac{5}{3}-\dfrac{1}{\log 2}$

The following sketch contains the graphs of $y=2^x$ and $y=\sqrt{x}+1$.

From the sketch, it looks like the two curves cross when $x=0$ and when $x=1$ and nowhere (To verify analytically that the curves have no other crossings, write $f(x)=\sqrt{x}+1-2^x$ and compute $f'(x)= \frac{1}{2\sqrt{x}}-(\log 2)2^x$. Notice that $f'(x)$ decreases as $x$ increases and so can take the value $0$ for at most a single value of $x$. Then, by the mean value theorem (or Rolle's theorem, which is Theorem 2.13.1 in the CLP-1 text), $f(x)$ can take the value $0$ for at most two distinct values of $x$.) else. Indeed, when $x=0$ we have $2^x=\sqrt{x}+1=1$ and when $x=1$ we have $2^x=\sqrt{x}+1=2$.

To antidifferentiate $2^x$, we write $2^x={(e^{\log 2})}^x=e^{x\log 2}$.

Draw a sketch first. You can also exploit a symmetry of the region to simplify your solution.

$\dfrac{8}{\pi}-1$

Here is a sketch of the specified region.

Both functions are even, so the region is symmetric about the $y$–axis. So, we will compute the area of the part with $x\ge 0$ and multiply by $2$. The curves $y=\sqrt{2} \cos(\pi x/4)$ and $y=x$ intersect when $x=\sqrt{2} \cos(\pi x/4)$ or $\cos(\pi x/4)=\frac{x}{\sqrt{2}}$, which is the case (The solution $x=1$ was found by guessing. To guess a solution to $\cos(\pi x/4)=\frac{x}{\sqrt{2}}$ just ask yourself what simple angle has a cosine that involves $\sqrt{2}$. This guessing strategy is essentially useless in the real world, but works great on problem sets and exams.) when $x=1$. So, using vertical strips as in the figure above, the area (including the multiplication by 2) is

Figure out where the two curves cross. To determine which curve is above the other, try evaluating $f(x)$ and $g(x)$ for some simple value of $x$. Alternatively, consider $x$ very close to zero.

$\dfrac{20}{9}$

For our computation, we will need an antiderivative of $x^2\sqrt{x^3+1}$, which can be found using the substitution $u=x^3+1$, $\dee{u} = 3x^2\,\dee{x}$:

The two functions $f(x)$ and $g(x)$ are clearly equal at $x=0$. If $x\ne0$, then the functions are equal when

The function $g(x)=3x^2$ is the larger of the two on the interval $[0,2]$, as can be seen by plugging in $x=1$, say, or by observing that when $x$ is very small $f(x)=x^2\sqrt{x^3+1}\approx x^2$ and $g(x)=3x^2$.

The area in question is therefore:

Think about whether it will easier to use vertical strips or horizontal strips.

$\dfrac{1}{6}$

First, let's figure out what our curve $x=y^2+y=y(y+1)$ looks like.

This leads to the figure below. We're evaluating the area from $y=-1$ to $y=0$. Since $y^2+y$ is negative there, the length of our (horizontal) slices are $0-(y^2+y)$.

Writing an integral for this is nasty. How can you avoid it?

$2\pi$

Let's begin by sketching our region. Note that $y=\sqrt{1-x^2}$ and $y=\sqrt{9-x^2}$ are the top halves of circles centred at the origin with radii 1 and 3, respectively.

Our region is the difference of two quarter-circles, so we find its area using geometry:

You are asked for the area, not the signed area. Be very careful about signs.

$2\Big[\pi-\frac{1}{4}\pi^2\Big]$

We will compute the area by using thin vertical strips, as in the sketch below:

By looking at the sketch above, we guess the line $y = 4 + 2\pi - 2x$ intersects the curve $y = 4 + \pi \sin x$ when $x=\frac{\pi}{2},$ $x=\pi$, and $x=\frac{3\pi}{2}$. Let's make sure these are correct by plugging them into the two equations, and making sure the $y$-values match:

Also from the sketch, we see that:

Now we can calculate:

You are asked for the area, not the signed area. Draw a sketch of the region and be very careful about signs.

$\dfrac{31}{6}$

First, here is a sketch of the region. We are not asked for it, but it is crucial for understanding the question.

The two curves $y=x+2$ and $y=x^2$ cross at $(2,4)$. The area of the part between them with $0\le x\le 2$ is:

The area of the part between the two curves with $2\le x\le 3$ is:

The total area is $\dfrac{10}{3}+\dfrac{11}{6}=\dfrac{31}{6}$.

You have to determine whether

One way to do so is to study the sign of $f(x)-g(x) = x\big(\sqrt{25-x^2}-3\big)$.

$\dfrac{26}{3}$

We need to figure out which curve is on top, when. To do this, set $h(x) = 3x - x\sqrt{25-x^2}$. If $h(x) > 0$, then $y=3x$ is the top curve; if $h(x)<0$, then $y=x\sqrt{25-x^2}$ is the top curve.

We only care about values of $x$ in $[0,4]$, so $x$ is nonnegative. Then $h(x)$ is positive when:

That is, $h(x)$ is never positive over the interval $[0,4]$. So, $y = x \sqrt{25-x^2}$ lies above $y=3x$ for all $0\le x\le 4$.

The area we need to calculate is therefore:

To evaluate $A_1$, we use the substitution $u(x) = 25-x^2$, for which $\dee{u} = u'(x)\,\dee{x}= -2x\,\dee{x}$; and $u(4)=25-4^2=9$ when $x=4$, while $u(0)=25-0^2=25$ when $x=0$. Therefore

For $A_2$ we use the antiderivative directly:

Therefore the total area is:

Flex those geometry muscles.

$\dfrac{7\pi}{8}-\dfrac{1}{2}$

Let's begin by sketching our region. Note that $y=\sqrt{9-x^2}$ is the top half of a circle centred at the origin with radius 3, while $y=\sqrt{1-(x-1)^2}$ is the top half of a circle of radius 1 centred at $(1,0)$.

Note $y=x$ intersects $y=\sqrt{1-(x-1)^2}$ at $(1,1)$, the highest part of the smaller half-circle.

We can easily take the area of triangles and sectors of circles. With that in mind, we cut up our region the following way:

So, the area of our region is $\dfrac{9\pi}{8} - \dfrac{1}{2}-\dfrac{\pi}{4}=\dfrac{7\pi}{8}-\dfrac{1}{2}$.

These two functions have three points of intersection. This question is slightly messy, but uses the same concepts we've been practicing so far.

$12\sqrt{2}-\dfrac{13}{4}$

The first function is a cubic, with intercepts at $x=0,\pm2$. The second is a straight line with a positive slope.

We need to figure out what these functions look like in relation to one another, so let's find their points of intersection.

So, our three points of intersection are when $x=2$ and when ${\color{red}x=-1\pm\sqrt{2}}$. We note

So, we need to see which function is on top over the two intervals $\left[-1-\sqrt{2},-1+\sqrt{2}\right]$ and $\left[-1+\sqrt{2},2\right]$. It suffices to check points in these intervals.

Since 0 is in the interval $\left[-1-\sqrt{2},-1+\sqrt{2}\right]$, $x(x^2-4)$ is the top function in that interval. Since 1 is in the interval $\left[-1+\sqrt{2},2\right]$, $x-2$ is the top function in that interval. Now we can set up the integral to evaluate the area:

After some taxing but rudimentary algebra: