please sove the attached question using integration method as I am not able to get how to use integration here.

To assist you effectively, I’ll guide you through solving a problem using integration. While I don’t have the specific question you’re referring to, I can illustrate the integration method with a common example. Let’s consider finding the area under a curve, which is a typical application of integration.

Understanding the Problem

Suppose we want to find the area under the curve of the function \( f(x) = x^2 \) from \( x = 1 \) to \( x = 3 \). This means we want to calculate the integral of \( f(x) \) over that interval.

Setting Up the Integral

The area \( A \) under the curve can be expressed as:

A = ∫ from 1 to 3 of \( f(x) \, dx \)

In our case, this translates to:

A = ∫ from 1 to 3 of \( x^2 \, dx \)

Calculating the Integral

To solve this integral, we first need to find the antiderivative of \( x^2 \). The antiderivative of \( x^n \) is given by:

\( \frac{x^{n+1}}{n+1} + C \)

For our function \( x^2 \), the antiderivative is:

\( \frac{x^{3}}{3} + C \)

Evaluating the Definite Integral

Now, we evaluate the definite integral from 1 to 3:

A = \( \left[ \frac{x^{3}}{3} \right]_{1}^{3} \)

This means we will calculate:

• First, substitute \( x = 3 \): \( \frac{3^{3}}{3} = \frac{27}{3} = 9 \)
• Next, substitute \( x = 1 \): \( \frac{1^{3}}{3} = \frac{1}{3} \)

Now, we subtract the two results:

A = 9 - \( \frac{1}{3} \)

To simplify, convert 9 into a fraction: \( 9 = \frac{27}{3} \)

Thus, A = \( \frac{27}{3} - \frac{1}{3} = \frac{26}{3} \)

Final Result

The area under the curve \( f(x) = x^2 \) from \( x = 1 \) to \( x = 3 \) is \( \frac{26}{3} \) square units. This method of integration allows us to find areas under curves, which is a fundamental application in calculus.

If you have a specific problem in mind, feel free to share it, and I can help you work through that particular question using integration!