Question

Use areas to evaluate the integral. integral_a^7b 2s ds, 0 < a< b integral_a^7b 2s ds =

Answer 1

To solve the integral from a to 7b of 2s ds, we apply the power rule of integration to find the indefinite integral s^2 + C and then evaluate it from a to 7b, resulting in the final answer 2((7b)^2 - a^2).

Step-by-step explanation:

To evaluate the integral of 2s ds from a to 7b, where 0 < a < b, we recognize that this is a straightforward integral with respects to s. The function 2s is a linear function, which when integrated from a to 7b, represents the area under the curve of the function over that interval. The integral in question is a basic rule of integration applying to polynomials.

The steps to solve the integral are:

1. Apply the power rule of integration, which states that ∫ x^n dx = x^(n+1)/(n+1) + C, where C is the constant of integration.
2. For the integral ∫ 2s ds, the function to be integrated is 2s, which can be considered as 2s^1.
3. Following the power rule, we integrate to get s^2 + C.
4. After finding the indefinite integral, we apply the limits of integration from a to 7b, resulting in (7b)^2 - a^2.
5. Multiply through by the constant 2 to obtain the final result: 2((7b)^2 - a^2).

Therefore, the evaluated integral equals 2((7b)^2 - a^2).