Area Between Curves Exercises

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Here we’ll practice finding area between curves.

Find the area of the region bounded by the curves $y = 1$ and $y =7+6x-x^2$ .

First we find the points of intersections. $\begin{align*} 7+6x-x^2 &= 1\\ x^2-6x-6 &= 0\\ (x-3)^2 - 15 &= 0\\ (x-3)^2 &= 15 \\ x-3 &= \pm \sqrt(15) \\ x &= 3 +\sqrt{15} \textrm{ or }3 - \sqrt{15} \end{align*}$

By graphing the two functions, we can see that $y =7+6x-x^2$ is always greater than $1$ on the interval $[3 - \sqrt {15}, 3 + \sqrt {15}]$ .

Thus the area of this region is $\int _{3-\sqrt {15}}^{3+\sqrt {15}} (7+6x-x^2) - 1 \d x$

$\begin{align*} \int_{3 - \sqrt{15}}^{3+\sqrt{15}} (7+6x-x^2) - 1 \d x &= \int_{3 - \sqrt{15}}^{3+\sqrt{15}} 6+6x-x^2 \d x\\ &= \eval{6x+3x^2-\frac{1}{3}x^3}_{3 - \sqrt{15}}^{3+\sqrt{15}}\\ &= 20\sqrt{15} \end{align*}$

$\textrm {Area} = \answer {20\sqrt {15}}$

Find the area of the region bounded by the curves $y = x$ and $y = \frac {1}{2} x + 1$ in the first quadrant.

First we find the point of intersections. $\begin{align*} x &= \frac{1}{2}x+1\\ \frac{1}{2}x &=1\\ x &=2 \end{align*}$

$y = \frac {1}{2}x+1$ is greater than $y=x$ on the interval $[0,2]$

Thus the area of this region is

$\int _0^2 (\frac {1}{2}x+1) - x \d x$

$\begin{align*} \int_0^2 (\frac{1}{2}x+1) - x \d x &= \int_0^2 (\frac-{1}{2}x+1) \d x\\ &= \eval{-\frac{1}{4}x^2+x}_0^2\\ &= 1 \end{align*}$

$\textrm {Area} = \answer {1}$

Since this region is a triangle with base of length $1$ on the $y$ -axis, and height $2$ , one could also have calculated this area geometrically.

Find the area of the region bounded by two consecutive intersections of the curves $y=\sin (x)$ and $y = \cos (x)$ .

We can use any two consecutive intersections, but the first two positive intersections are convenient.

These occur at $x = \frac {\pi }{4}$ and $x = \frac {5\pi }{4}$

$y = \sin (x)$ is greater than $y=\cos (x)$ on the interval $[\frac {\pi }{4},\frac {5\pi }{4}]$

Thus the area of this region is $\int _{\frac {\pi }{4}}^{\frac {5\pi }{4}} \sin (x) - \cos (x) \d x$

$\begin{align*} \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \sin(x) - \cos(x) \d x &= -\eval{\cos(x)+\sin(x)}_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \d x\\ &=-(-\frac{2}{\sqrt{2}} - \frac{2}{\sqrt{2}})\\ &= \frac{4}{\sqrt{2}} \end{align*}$

$\textrm {Area} = \answer {\frac {4}{\sqrt {2}}}$

Find the area of the region bounded by the vertical lines $x=0$ and $x=1$ , and the curves $y = x^2+1$ and $y = x^3$ .

There are no intersection points between the curves on this interval, and $y = x^2+1$ is always above $y=x^3$ on this interval.

Thus the area of this region is

$\int _{0}^{1} (x^2+1)-x^3 \d x$

$\begin{align*} \int_{0}^{1} (x^2+1)-x^3 \d x &= \eval{\frac{x^3}{3}+x-\frac{x^4}{4}}_0^1 \d x\\ &= \frac{13}{12} \end{align*}$

$\textrm {Area} = \answer {\frac {13}{12}}$

Find the area of the region bounded by the curves $y=x-2$ and $y=-x$ , and $y = 4+2x-x^2$ .

First sketch a picture of the region

Next we find the $x$ -coordinates of the relevant points of intersection.

Let $a$ be the point of intersection between $y = -x$ and $y = 4+2x-x^2$ , $b$ between $y=-x$ and $y=x-2$ , and $c$ between $y=x-2$ and $y=4+2x-x^2$ .

Then

$\begin {cases} -a = 4+2a-a^2\\ -b = b-2\\ c-2 = 4+2c-c^2 \end {cases}$

$\begin {cases} 0 = 4+3a-a^2\\ 0 = 2b-2\\ 0= 6+c-c^2 \end {cases}$

$\begin {cases} 0 = (1+a)(4-a)\\ b = 1\\ 0= (3-c)(2+c) \end {cases}$

Thus $a=-1$ , $b = 1$ , and $c = 3$ (make sure you understand why $a=4$ and $c=-2$ are not the correct intersections).

Thus the area of this region is

$\int _{-1}^{1} (4+2x-x^2) - (-x) \d x + \int _{1}^{3} (4+2x-x^2) - (x-2) \d x$

The first summand is $\begin{align*} \int_{-1}^{1} (4+2x-x^2) - (-x) \d x &= \int_{-1}^{1} (4+3x-x^2) \d x\\ &= \eval{4x+\frac{3}{2}x^2-\frac{1}{3}x^3}_{-1}^1\\ &= \frac{22}{3} \end{align*}$

The second summand is

$\begin{align*} \int_{1}^{3} (4+2x-x^2) - (x-2) \d x &= \int_{1}^{3} (6+x-x^2) \d x\\ &= \eval{6x+\frac{x^2}{2}-\frac{x^3}{3}}_1^3\\ &= \frac{22}{3} \end{align*}$

Note that these two area equal, which we could have also discovered by observing the symmetry of the figure about the line $x=1$ .

$\textrm {Area} = \answer {\frac {44}{3}}$

Find the area of the region bounded by the curves $y=\sqrt [3]{x}$ and $y=\frac {2}{7}(x-1)$ and above the $x$ -axis. Try to solve this problem using horizontal rectangles, i.e. by integration with respect to $y$ .

First sketch a picture of the region

Rather than using two different regions of integration, this problem is a great case opportunity to use horizontal rectangles.

Solving for $x$ , we can rewrite the curves as $x = y^3$ and $x = \frac {7y}{2}+1$ .

The curves intersect at the point $(8,2)$

The area of this region is

$\int _{0}^{2} ( \frac {7y}{2}+1)-y^3 \d y$

$\begin{align*} \int_{0}^{2} ( \frac{7y}{2}+1)-y^3 \d y &= \eval{\frac{7y^2}{4}+y-\frac{y^4}{4}}_0^2\\ &= 7+2-4\\ &= 5 \end{align*}$

$\textrm {Area} = \answer {5}$

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