Question

Find the area of the region bounded by the given curves.

\(y = 1/x,\;\;\;y = {x^2},\;\;\;y = 0,\;\;\;x = e\)

Short answer

The area of the shaded region is \(\frac{{{e^3} - 4}}{3}\)

Step-by-step solution

Given information

The given value is:

\(y = 1/x,\;\;\;y = {x^2},\;\;\;y = 0,\;\;\;x = e\)

Find the points of intersection of the curves:

First, let us determine the intersections of the curves to find integral limits

\(\left\{ {\begin{aligned}{{c}}{y = \frac{1}{x}}\\{y = {x^2}}\end{aligned}} \right.\)

Set the sides equal

\(\frac{1}{x} = {x^2}\)

Solve the equation

\(x = 1\)

Substitute the value of\(x\)

\(y = 1\)

A possible solution is

\((x,y) = (1,1)\)

Find the x-intercept

The area of the region bounded by \(f(x) = {x^2}\)and \(g(x) = \frac{1}{x}\)in \(1 \le x \le e\) is:

\(A = \int_1^e {\left( {{x^2} - \frac{1}{x}} \right)} \)

Evaluate the area A

\(\int_1^e {{x^2} - \frac{1}{x}dx} \)

Isolate the function

\(f(x) = {x^2} - \frac{1}{x}\)

Find the x-intercepts

\(\begin{aligned}{}f(x) &= 0\\0 &= {x^2} - \frac{1}{x}\\0 &= {x^2} - \frac{1}{x},x \ne 0\\\frac{{{x^3} - 1}}{x} &= 0\end{aligned}\)

\(\begin{aligned}{}{x^3} - 1 &= 0\\x &= 1\end{aligned}\)

The total area is the absolute value of the initial integral

The graph

Evaluate the integral

\(A = \int_1^e {{x^2} - \frac{1}{x}dx} \)

\(\int {{x^2} - \frac{1}{x}dx} \)

Use the property of integral

\(\int {f(x) \pm g(x)dx = \int {f(x)dx \pm \int {g(x)dx} } } \)

\(\int {{x^2}dx - \int {\frac{1}{x}dx} } \)

Use \(\int {{x^n}dx} = \frac{{{x^{n + 1}}}}{{n + 1}},n \ne - 1\) to evaluate the integral

Use \(\int {\frac{1}{x}dx = In\left( {\left| x \right|} \right)} \) to evaluate the integral

\(\frac{{{x^3}}}{3} - In\left( {\left| x \right|} \right)\)

To evalute the define integral, return the limits of integration

\(\left( {\frac{{{x^3}}}{3} - In\left( {\left| x \right|} \right)} \right)_1^e\)

Evaluate the expression

\(\frac{{{e^3}}}{3} - In\left( {\left| e \right|} \right) - \left( {\frac{{{1^3}}}{3} - In\left( {\left| 1 \right|} \right)} \right)\)

\(A = \frac{{{e^3} - 4}}{3}\)