Question

In Exercises \(27-40\) , evaluate each integral using Part 2 of the Fundamental Theorem. Support your answer with NINT if you are unsure. $$\int_{-1}^{1}(r+1)^{2} d r$$

Short answer

The value of the integral is 8/3

Step-by-step solution

Setting up the Integral

The integral can be written as \(\int_{-1}^{1}(r+1)^{2} dr\), which is a definite integral over the interval of \(-1\) to \(1\) of the function \((r+1)^{2}\).

Find the Antiderivative

To evaluate this integral, first need to find the antiderivative F(r) of the function \((r+1)^{2}\). The antiderivative of \((r+1)^{2}\) is \(\frac{1}{3}(r+1)^{3}\) using the power rule.

Apply the Fundamental Theorem

The Fundamental Theorem of Calculus, Part 2, states that the definite integral of a function from a to b is equal to the antiderivative at b minus the antiderivative at a. So \(\int_{-1}^{1}(r+1)^{2} dr = \(\frac{1}{3}(1+1)^{3} - \frac{1}{3}((-1)+1)^{3}\)

Evaluate the Function

Now, it is possible to evaluate the integral by plugging in the respective values into the function: \(\frac{1}{3}(1+1)^{3}\) gives 8/3, \(\frac{1}{3}((-1)+1)^{3}\) gives 0. Subtracting these two results gives the value of the integral: 8/3 - 0 = 8/3