# Find the area bounded by the curves y=2 \sqrt{x} \ and \ y=x-5 between x=4 and __5__. So...

## Question:

Find the area bounded by the curves {eq}y=2 \sqrt{x} \ and \ y=x-5 {/eq} between {eq}x=4 {/eq}

and 5. So area=??

Find the area bounded by the t-axis (seconds) and {eq}y(t)=6sin( \frac{t}{18}) {/eq} between t=3 and 8. area=??

## Area Using Calculus

Calculus can be used to find the area {eq}A {/eq} between the graphs of two function {eq}y=f(x) {/eq} and {eq}y=g(x) {/eq}. The method is to first find where the graphs intersect by solving {eq}f(x)=g(x) {/eq} for {eq}x {/eq}. Then determine which graph {eq}y=f(x) {/eq} or {eq}y=g(x) {/eq} is above the other graph on one or both sides of a point of intersection. In this particular question no point of intersection will be in an interval over which the area {eq}A {/eq} is being calculated. After determining which graph is above the other the area can be calculated using calculus. If, for example, the graph {eq}y=f(x) {/eq} is above the graph {eq}y=g(x) {/eq} between {eq}x=a {/eq} and {eq}x=b {/eq} then the area {eq}A {/eq} is calculated as {eq}A=\int_{x=a}^b (f(x)-g(x))~dx {/eq}. Similar remarks hold for a function {eq}y=f(t) {/eq} where the {eq}t {/eq}-axis replaces the {eq}x {/eq}-axis.

The question is restated with slightly different notation. Find the area bounded by the

(a) curves {eq}y=2\sqrt{x} {/eq} and {eq}y=x-5 {/eq} between {eq}x=4 {/eq} and {eq}x=5 {/eq}.

(b) {eq}t {/eq}-axis (seconds) and {eq}y(t)=6\sin(\frac{t}{18}) {/eq} between {eq}t=3 {/eq} and {eq}t=8 {/eq}.

Consider part (a). Determine where the curves intersect by solving

{eq}\begin{eqnarray*} 2\sqrt{x} &=& x-5 \\ 4x &=& (x-5)^2 \\ 4x &=& x^2-10x+25 \\ 0 &=& x^2-14x+25. \end{eqnarray*} {/eq}

Use the quadratic formula to solve the equation to get

{eq}\begin{eqnarray*} x &=& \frac{-(-14)\pm\sqrt{(-14)^2-4(1)25}}{2(1)} \\ &=& \frac{14\pm\sqrt{96}}{2} \\ &=& \frac{14\pm4\sqrt{6}}{2} \\ &=& 7\pm2\sqrt{6}. \end{eqnarray*} {/eq}

A quick check shows that only {eq}x=7+2\sqrt{6} {/eq} satisfies the original equation. Since {eq}x=7+2\sqrt{6}\doteq11.9 {/eq} is not in the interval {eq}\left[4,5\right] {/eq} then the graphs do not cross in the interval where the area is being calculated. Hence, one curve remains above the other in the interval {eq}\left[4,5\right] {/eq}. To determine which curve is above the other pick a point in the interval {eq}\left[4,5\right] {/eq}, say {eq}x=4.5 {/eq} and plug it into the equation for each graph to get

{eq}\begin{eqnarray*} y &=& 2\sqrt{4.5} \\ &\doteq&4.2 \\ y &=& 4.5-5 \\ &=& -0.5. \end{eqnarray*} {/eq}

Consequently, the curve {eq}y=2\sqrt{x} {/eq} is above {eq}y=x-5 {/eq} in the interval {eq}\left[4,5\right] {/eq}. Hence, the area {eq}A {/eq} bounded by the curves is

{eq}\begin{eqnarray*} A &=& \int_{x=4}^5 (2\sqrt{x}-(x-5))~dx \\ &=& \int_{x=4}^5 (2\sqrt{x}-x+5)~dx \\ &=& \left.\frac{4x^\frac{3}{2}}{3}-\frac{x^2}{2}+5x\right|_{x=4}^5 \\ &=& \frac{4(\sqrt{5})^3}{3}-\frac{25}{2}+25-\left(\frac{4(\sqrt{4})^3}{3}-\frac{16}{2}+20\right) \\ &=& \frac{40\sqrt{5}-61}{6}. \end{eqnarray*} {/eq}

Consider part (b). It is easy to see that {eq}y(t)=6\sin\left(\frac{t}{18}\right) {/eq} is above the {eq}t {/eq}-axis for {eq}3\leq t\leq8 {/eq}. Consequently, the area {eq}A {/eq} is

{eq}\begin{eqnarray*} A &=& \int_{t=3}^8 6\sin\left(\frac{t}{18}\right)~dt \\ &=& -108\int_{t=3}^8 -\frac{1}{18}\sin\left(\frac{t}{18}\right)~dt \\ &=& -108\left(\left.\cos\left(\frac{t}{18}\right)\right|_{t=3}^8\right) \\ &=& -108\left(\cos\left(\frac{4}{9}\right)-\cos\left(\frac{1}{6}\right)\right) \\ &=& 108\left(\cos\left(\frac{1}{6}\right)-\cos\left(\frac{4}{9}\right)\right) \\ &\doteq& 9 \end{eqnarray*} {/eq}

rounded to the nearest whole number. 