QuestionJuly 14, 2025

Fill in the Blank 1 point Find the area bound by y=(1)/(x^2) and the x-axis over the interval [1,5] Area=boxed (type your answer... units^2)

Fill in the Blank 1 point Find the area bound by y=(1)/(x^2) and the x-axis over the interval [1,5] Area=boxed (type your answer... units^2)
Fill in the Blank 1 point Find the area bound by y=(1)/(x^2) and the x-axis over the interval [1,5] Area=boxed (type your answer... units^2)

Solution
4.1(340 votes)

Answer

\frac{4}{5} \text{ units}^2 Explanation 1. Set up the integral The area under the curve y = \frac{1}{x^2} from x=1 to x=5 is given by the definite integral \int_{1}^{5} \frac{1}{x^2} \, dx. 2. Integrate the function The antiderivative of \frac{1}{x^2} is -\frac{1}{x}. So, \int \frac{1}{x^2} \, dx = -\frac{1}{x} + C. 3. Evaluate the definite integral Evaluate -\frac{1}{x} from 1 to 5: -\frac{1}{5} - (-\frac{1}{1}) = -\frac{1}{5} + 1 = \frac{4}{5}.

Explanation

1. Set up the integral<br /> The area under the curve $y = \frac{1}{x^2}$ from $x=1$ to $x=5$ is given by the definite integral $\int_{1}^{5} \frac{1}{x^2} \, dx$.<br />2. Integrate the function<br /> The antiderivative of $\frac{1}{x^2}$ is $-\frac{1}{x}$. So, $\int \frac{1}{x^2} \, dx = -\frac{1}{x} + C$.<br />3. Evaluate the definite integral<br /> Evaluate $-\frac{1}{x}$ from 1 to 5: <br /> $-\frac{1}{5} - (-\frac{1}{1}) = -\frac{1}{5} + 1 = \frac{4}{5}$.
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