Question

Consider the region satisfying the inequalities. Find the area of the region. $$ y \leq \frac{1}{x^{2}}, y \geq 0, x \geq 1 $$

Short answer

The area of the region is 1.

Step-by-step solution

Identify the region

The region is defined by the inequalities, \( y \leq \frac{1}{x^{2}} \), \( y \geq 0 \), \( x \geq 1 \). This is the region bounded by the curve \( y = \frac{1}{x^{2}} \) and the lines \( x = 1 \) and the x-axis.

Calculate the Definite Integral of the function

The area under the curve can be calculated by the definite integral of the function \( y = \frac{1}{x^{2}} \), from the lower limit of x which is 1, to the upper limit of x, which is infinity. So, the integral becomes \(\int_1^{\infty} \frac{1}{x^{2}} \, dx \).

Solve the Definite Integral

Solving the integral, which is \(\int_1^{\infty} \frac{1}{x^{2}} \, dx \), the integral of \( \frac{1}{x^{2}} \) is \( -\frac{1}{x} \). So, evaluate it at infinity and 1, we have \( -\frac{1}{\infty} + 1 \).

Get the Area

Using the property that any real number divided by infinity is zero, the final answer will be \( 0 + 1 = 1 \).