Chapter 5.5, Problem 61E

To determine

To find: The equal are of the equation from the following equation:

$y=e^{\sqrt{x}}$, $y=2x{e}^x$, and $y=e^{\sin x}\sin 2x$

Answer to Problem 61E

All the equations of the areas are equal.

Explanation of Solution

Given:

The equations are

$y=e^{\sqrt{x}},\left(0,1\right)$.

$y=2x{e}^x,\left(0,1\right)$

$y=e^{\sin x}\sin 2x,\left(0,\frac{\pi }{2}\right)$

Calculation:

Express the area of function by using formula:

$A={\displaystyle {\int }_a^{b}ydx}$ (1)

The equation is,

$y=e^{\sqrt{x}}$

Calculate the area of the equation:

Substitute 0 for a, 1 for b, and $e^{\sqrt{x}}$ for y in Equation (1).

$A_1={\displaystyle {\int }_0^1{e}^{\sqrt{x}}dx}$ (2)

Consider $u=\sqrt{x}$ (3)

Differentiate both sides of the Equation (3).

$\begin{array}{c}du=\frac{1}{2\sqrt{x}}dx\\ 2\sqrt{x}du=dx\end{array}$ (4)

Substitute u for $\sqrt{x}$ in Equation (4).

$2udu=dx$

Calculate the lower limit value of u using Equation (3).

Substitute 0 for x in Equation (3).

$\begin{array}{c}u=\sqrt{0}\\ =0\end{array}$

Calculate the upper limit value of u using Equation (3).

Substitute 1 for x in Equation (3).

$\begin{array}{c}u=\sqrt{1}\\ =1\end{array}$

Apply lower and upper limits for u in Equation (2).

Substitute u for $\left(\sqrt{x}\right)$ and $\left(2udu\right)$ for dx in Equation (2).

$\begin{array}{c}{A}_1={\displaystyle {\int }_0^1{e}^u\left(2udu\right)}\\ =2{\displaystyle {\int }_0^1{e}^{u}udu}\end{array}$ (5)

Integrate Equation (5).

$\begin{array}{c}{A}_1=2{\left[u{e}^u-{\displaystyle \int 1.e^u}\right]}_0^1\\ =2{\left[u{e}^u-e^u\right]}_0^1\\ =2\left[\left(1\times e^1-e^1\right)-\left(0\times e^0-e^0\right)\right]\end{array}$

$\begin{array}{c}=2\left(0+1\right)\\ =2\end{array}$

The equation is

$y=2x{e}^x$

Calculate the area of the equation:

Substitute 0 for a, 1 for b, and $2x{e}^x$ for y in Equation (1).

$A_2={\displaystyle {\int }_0^{1}2x{e}^{x}dx}$ (6)

Integrate Equation (6).

$\begin{array}{c}{A}_2=2{\left[x{e}^x-{\displaystyle \int 1.e^x}\right]}_0^1\\ =2{\left[x{e}^x-e^x\right]}_0^1\\ =2\left[\left(1\times e^1-e^1\right)-\left(0\times e^0-e^0\right)\right]\end{array}$

$\begin{array}{c}=2\left(0+1\right)\\ =2\end{array}$

The equation is,

$y=e^{\sin x}\sin 2x$

Calculate the area of the equation:

Substitute 0 for a, $\frac{\pi }{2}$ for b, and $e^{\sin x}\sin 2x$ for y in Equation (1).

$\begin{array}{c}{A}_3={\displaystyle {\int }_0^{\frac{\pi }{2}}{e}^{\sin x}\sin 2xdx}\\ ={\displaystyle {\int }_0^{\frac{\pi }{2}}{e}^{\sin x}\left(2\sin x\cos x\right)dx}\\ =2{\displaystyle {\int }_0^{\frac{\pi }{2}}{e}^{\sin x}\sin x\cos xdx}\end{array}$ (7)

Consider $u=\sin x$ (8)

Differentiate both sides of the Equation (8).

$du=\cos xdx$

Calculate the lower limit value of u using Equation (8).

Substitute 0 for x in Equation (8).

$\begin{array}{c}u=\sin 0\\ =0\end{array}$

Calculate the upper limit value of u using Equation (8).

Substitute $\frac{\pi }{2}$ for x in Equation (8).

$\begin{array}{c}u=\sin \left(\frac{\pi }{2}\right)\\ =1\end{array}$

Apply lower and upper limits for u in Equation (7).

Substitute u for $\left(\sin x\right)$ and $du$ for $\cos xdx$dx in Equation (7).

$A_3=2{\displaystyle {\int }_0^1{e}^{u}udu}$ (9)

Integrate Equation (9).

$\begin{array}{c}{A}_3=2{\left[u{e}^u-{\displaystyle \int 1.e^u}\right]}_0^1\\ =2{\left[u{e}^u-e^u\right]}_0^1\\ =2\left[\left(1e^1-e^1\right)-\left(0\times e^0-e^0\right)\right]\end{array}$

$\begin{array}{c}=2\left(0+1\right)\\ =2\end{array}$

The three areas are equal,

Therefore, all the equations of the areas are equal.