Problem 27 - Calculus Early Transcendentals

Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int(5 s+3)^{2} d s$$

Step-by-Step Solution

Answer

Question: Find the indefinite integral of the function $(5s+3)^2$, then verify the result by taking the derivative of the obtained function. Answer: The indefinite integral of the given function is $\frac{25}{3}s^3 + 15s^2 + 9s + C$. The derivative of this function is $25s^2 + 30s + 9$, which matches the original function inside the integral, confirming that the integration was performed correctly.

Step 1: Expand the integrand

Expand the given function inside the integral: $$(5s + 3)^2 = (5s)^2 + 2(5s)(3) + 3^2 = 25s^2 + 30s + 9$$ So, the given integral becomes: $$\int (5s + 3)^2 ds = \int (25s^2 + 30s + 9) ds$$

Step 2: Integrate term by term

Now, integrate each term of the expanded function: $$\int (25s^2 + 30s + 9) ds = 25\int s^2 ds + 30\int s ds + 9\int 1 ds$$ Recalling that the integral of \(s^n\) is given by \(\frac{s^{n+1}}{n+1}\) and the integral of a constant is given by the constant times the variable, we have: $$25\int s^2 ds + 30\int s ds + 9\int 1 ds = 25\frac{s^3}{3} + 30\frac{s^2}{2} + 9s + C$$ which simplifies to: $$\frac{25}{3}s^3 + 15s^2 + 9s + C$$

Step 3: Differentiate to check

Now, differentiate the obtained function to check if we get the original function inside the integral back: $$\frac{d}{ds} \left(\frac{25}{3}s^3 + 15s^2 + 9s + C\right) = \frac{75}{3}s^2 + 30s + 9 = 25s^2 + 30s + 9$$ Since the derivative of our obtained function matches the original function inside the integral, our integration was performed correctly. Thus, the indefinite integral of the given function is: $$\int (5s+3)^2 ds = \frac{25}{3}s^3 + 15s^2 + 9s + C$$