Application Of Integrals

We are given the integral: We can simplify this by performing substitution. Let , so that and . The integral becomes: Simplifying further: This can be split into two parts: The first part gives , and the second part involves integration by parts or recognizing standard forms, resulting in:

We are given that . To find , we need to integrate with respect to . First, break it into parts: Now integrate each term separately: - The integral of is . - The integral of is . - The integral of 5 is . Thus, we have: We are given that . Substituting : Simplifying: Thus, the function is: The closest option is (3).

The given curve is , which represents a parabola. We need to find the area enclosed by the curve between and . First, solve for : To find the area, we integrate the function from to : The integral of is: Evaluating this from 0 to 1: Thus, the area is 2 square units.

We are asked to evaluate the integral: We will use the symmetry of the integral to simplify the calculation. First, make the substitution . This gives: Thus, the integral becomes: Now, rewrite the integral: This simplifies to: Thus, solving for : The integral is a standard result, which is . Thus:

We are asked to evaluate the integral: To solve this, we will use partial fraction decomposition. First, express the integrand as a sum of simpler fractions: Multiply both sides by the denominator : Now expand both sides: Simplify: Collect terms in powers of : Now equate the coefficients of like powers of : 1. Coefficient of : 2. Coefficient of : 3. Constant term: Solve this system of equations: From , we get . Substitute into the second equation: Substitute and into the third equation: Now substitute into and : Thus, the partial fraction decomposition is: Now integrate each term: The integrals are straightforward: