Problem 38 - Calculus Early Transcendentals

Question

Evaluate the following integrals or state that they diverge. $$\int_{-3}^{1} \frac{d x}{(2 x+6)^{2 / 3}}$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral: $$\int_{-3}^{1} \frac{d x}{(2 x+6)^{2 / 3}}$$ Answer: The definite integral evaluates to 3.

Step 1: Simplify the given function

First, we can simplify the given function by factoring out the constant from the denominator: $$\int_{-3}^{1} \frac{d x}{(2 x+6)^{2 / 3}} = \int_{-3}^{1} \frac{d x}{(2(x+3))^{2 / 3}}$$

Step 2: Rewrite the function

Next, we can rewrite the function as follows: $$\int_{-3}^{1} \frac{d x}{(2(x+3))^{2 / 3}} = \frac{1}{2^{2/3}} \int_{-3}^{1} (x+3)^{-2 / 3} dx$$

Step 3: Find the antiderivative of the function

Now we need to find the antiderivative of the function \((x+3)^{-2/3}\). Applying the power rule for integration: $$\int (x+3)^{-2/3} dx = \frac{(x+3)^{1/3}}{1/3} + C = 3(x+3)^{1/3} + C$$

Step 4: Apply the Fundamental Theorem of Calculus

Now that we have the antiderivative, we can apply the Fundamental Theorem of Calculus to find the value of the definite integral: $$\frac{1}{2^{2/3}} \int_{-3}^{1} (x+3)^{-2 / 3} dx = \frac{1}{2^{2/3}} \left[3(x+3)^{1/3}\right]_{-3}^{1}$$

Step 5: Evaluate the definite integral

Finally, we evaluate the definite integral by plugging the limits of integration into the antiderivative: $$\frac{1}{2^{2/3}} \left[3(x+3)^{1/3}\right]_{-3}^{1} = \frac{1}{2^{2/3}} \left[3(4)^{1/3} - 3(0)^{1/3}\right] = \frac{1}{2^{2/3}}[3(2^{\frac{2}{3}})] = 3$$ Thus, the value of the given integral is: $$\int_{-3}^{1} \frac{d x}{(2 x+6)^{2 / 3}} = 3$$