Question

Find the area of the region in the first quadrant bounded by the line \(y=x,\) the line \(x=2,\) the curve \(y=1 / x^{2},\) and the \(x\) -axis.

Short answer

The area of the region in the first quadrant bounded by the line \(y=x,\), the line \(x=2,\), the curve \(y=1 / x^{2},\) and the \(x\)-axis is 1 square unit.

Step-by-step solution

Identify the Bounding Curves and Areas

First, identify the different areas that are enclosed by the given functions. The area of concern is the one enclosed by the line \(y=x\), the curve \(y=1 / x^{2}\), the line \(x=2\) and the \(x\)-axis. Draw it out if necessary for better understanding.

Set the Integral Limits

The limits of integration for the area would be from 1 to 2 since that's where the intersection points are. You can ascertain this by graphing the functions or finding the intersection points mathematically. Substituting \(y=x\) into \(y=1 / x^{2}\), yields \(x^{3}=1\), whose positive root is 1 (since we are in the first quadrant). And the vertical line \(x=2\) is the other limit.

Write the Definite Integral Expression Representing the Area

The final step is to calculate the definite integral, that is, the difference of the two curves over the interval [1,2]. The area \(A\) of the region bounded by the curves y=f(x) and \(y=g(x)\), and the lines \(x=a\) and \(x=b\) is given by, \(A=\int_{a}^{b}[f(x)-g(x)] dx\). Here, \(f(x)=x\), and \(g(x)=frac{1}{x^{2}}\). So, the area can be defined as \(A=\int_{1}^{2}[x-\frac{1}{x^{2}}] dx\).

Evaluate the Definite Integral

To evaluate the integral, calculate the antiderivatives of the functions and apply the Fundamental Theorem of Calculus. The antiderivative of \(x\) is \(\frac{x^{2}}{2}\) and of \(1 / x^{2}\) is \(-1/x\). Plugging the limits into the antiderivatives gives, \(A=[\frac{x^{2}}{2}+1/x]|_{1}^{2} = [(2^{2}/2+1/2)-(1^{2}/2+1)] = 2 + 0.5 - 0.5 - 1 = 1.\)