Problem 24 - Calculus Early Transcendentals

Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int\left(3 u^{-2}-4 u^{2}+1\right) d u$$

Step-by-Step Solution

Answer

Question: Find the indefinite integral of the function 3u^{-2} - 4u^2 + 1, and check the solution by differentiation. Answer: The indefinite integral of the function is given by: -3u^{-1} - (4u^3)/3 + u + C, where C is the constant of integration. The solution is verified by differentiation which yields the original function: 3u^{-2} - 4u^2 + 1.

Step 1: Integrate each term separately

We will integrate 3 u^{-2}, -4 u^{2}, and 1 with respect to u. Recall the power rule for integration: $$\int x^n dx = \frac{x^{n+1}}{n+1} + C$$ Applying the power rule to each term: $$\int 3u^{-2} du = \frac{3u^{-1}}{-1} + C_1$$ $$\int -4u^2 du = \frac{-4u^3}{3} + C_2$$ $$\int 1 du = u + C_3$$

Step 2: Combine the integration results

Now that we have integrated each term separately, we can combine the results to obtain the general solution of the given indefinite integral $$\int(3u^{-2} - 4u^{2} + 1) du = \frac{3u^{-1}}{-1} + \frac{-4u^3}{3} + u + C$$ Where C is the constant of integration, and C = C_1 + C_2 + C_3. So our solution is: $$\int(3u^{-2} - 4u^{2} + 1) du = -3u^{-1} -\frac{4u^3}{3} + u + C$$

Step 3: Check the solution by differentiation

To verify our solution, we will differentiate it and ensure it matches the original function. Using the power rule for differentiation: $$\frac{d}{du}(-3u^{-1} -\frac{4u^3}{3} + u + C) = -3\cdot(-1)u^{-2} -4u^{2} + 1$$ This simplifies to: $$3u^{-2} - 4u^2 + 1$$ Since this matches the original function, we have demonstrated that our integration is correct.