Question

Find the area between the graph of ….and the r-axis on the interval [9, 16]. Write the exact answer. Do not round.

Answer 1

The area under a curve between two points can be found by doing a definite integral between the two points.

To find the area between the curve y = f(x) between x = a and x = b, integrate y = f(x) between the limits of a and b.

Given the function;

$f(x)=7\sqrt[]{x}$

and the x-interval is;

$\lbrack9,16\rbrack$

Thus, the area A between the graph and the x-interval is;

$A=\int ^{16}_97\sqrt[]{x}dx$

Next, we evaluate the integral, we have;

$\begin{gathered} \int 7\sqrt[]{x}dx=\int 7x^{\frac{1}{2}}dx \\ \int 7\sqrt[]{x}dx=\frac{7x^{\frac{1}{2}+1}}{\frac{1}{2}+1} \\ \int 7\sqrt[]{x}dx=\frac{7x^{\frac{3}{2}}}{\frac{3}{2}} \\ \int 7\sqrt[]{x}dx=\frac{14x^{\frac{3}{2}}}{3}+c \\ \text{Where c is the integral constant} \end{gathered}$

Then, we should apply the integral limits, we have;

$\begin{gathered} \int ^{16}_97\sqrt[]{x}dx=\lbrack\frac{14x^{\frac{3}{2}}}{3}\rbrack^{16}_9 \\ \int ^{16}_97\sqrt[]{x}dx=(\frac{14}{3}(16)^{\frac{3}{2}})-(\frac{14}{3}(9)^{\frac{3}{2}}) \\ \int ^{16}_97\sqrt[]{x}dx=\frac{14}{3}(64-27) \\ \int ^{16}_97\sqrt[]{x}dx=\frac{14}{3}(37) \\ \int ^{16}_97\sqrt[]{x}dx=\frac{518}{3} \end{gathered}$

Thus, the area A between the graph and the x-interval is;

$A=\frac{518}{3}\text{square units}$