Question

Graph the region between the curves and use your calculator to compute the area correct to five decimal places.

\(y = {\tan ^2}x,\quad y = \sqrt x \)

Short answer

The area of the region is 0.25141.

Step-by-step solution

The area \(A\) of the region bounded by the curves is given by

The area\(A\)of the region bounded by the curves\(y = f(x),y = g(x)\), and the lines\(x = a,x = b\), where\(f\)and\(g\)are continuous and\(f(x) \ge g(x)\)for all\(x\)in [a, b], is

Sketch the graph of the curves

Procedure to sketch the region bounded by the two curves is explained below:

- Substitute different values for \(x\)and draw the graph for the function \(y = {\tan ^2}x\).

- Substitute different values for \(x\)and plot for the function \(y = \sqrt x \).

- Shade the region where the points intersect on the curves.

The region enclosed by the curves \(y = {\tan ^2}x\) and \(y = \sqrt x \) is shown in below.

Calculate the area of the region enclosed.

The curves intersect at \(x = 0\)and \(x = 0.75\).

The curves are bounded by the top and bottom curve. Hence, so the integration can be done with respect to x.

Find the area of the region bounded by the curves using the relation:

\(A = \int_a^b {(f(} x) - g(x))dx(1)\)

Here, the top curve is \(f(x)\), the bottom curve is \(g(x)\), the lower limit is \(a\), and the upper limit is.

Substitute \(\sqrt x \) for \(f(x),{\tan ^2}x\) for \(g(x),0\) for \(a\), and 0.75 for \(b\) in Equation (1).

\(\begin{array}{l}A = \int_0^{0.75} {\left[ {\sqrt x - {{\tan }^2}x} \right]} dx\\A = \int_0^{0.75} {\left[ {{x^{\frac{1}{2}}} - \left( {{{\sec }^2}x - 1} \right)} \right]} dx\\A = \left[ {\frac{{{x^{\frac{1}{2} + 1}}}}{{\frac{1}{2} + 1}} - \tan x + x} \right]_0^{0.75}\\A = \left[ {\frac{2}{3}{x^{\frac{3}{2}}} - \tan x + x} \right]_0^{0.75}\end{array}\)

Solve the problem further

\(\begin{array}{l}A = \left( {\frac{2}{3}{{(0.75)}^{\frac{3}{2}}} - \tan (0.75) + 0.75} \right) - (0)\\A = 0.43301 - 0.93160 + 0.75\\A = 0.25141\end{array}\)