Problem 26 - Calculus Early Transcendentals

Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int\left(\frac{5}{t^{2}}+4 t^{2}\right) d t$$

Step-by-Step Solution

Answer

Question: Determine the indefinite integral of the function: $$\int\left(\frac{5}{t^{2}}+4 t^{2}\right) dt$$ Answer: The indefinite integral of the given function is: $$\int\left(\frac{5}{t^{2}}+4 t^{2}\right) dt = -\frac{5}{t} + \frac{4}{3} t^3 + C$$

Step 1: Break down the given integral into two separate integrals

The given integral can be separated as follows: $$\int\left(\frac{5}{t^{2}}+4 t^{2}\right) d t = \int \frac{5}{t^{2}} dt + \int 4 t^{2} dt$$

Step 2: Apply integration rules

Next, we will apply the integration rules for each integral, remembering to add the integration constant (C) at the end of each integral. For the first integral, we have: $$\int \frac{5}{t^{2}} dt = 5 \int \frac{1}{t^{2}} dt$$ We can rewrite this as: $$5 \int t^{-2} dt$$ Now integrate to get: $$5 \frac{t^{-1}}{-1} + C_1 = -\frac{5}{t} + C_1$$ For the second integral, we have: $$\int 4t^2 dt$$ Now integrate to get: $$4 \frac{t^3}{3} + C_2 = \frac{4}{3} t^3 + C_2$$ Combine both integrals: $$\int\left(\frac{5}{t^{2}}+4 t^{2}\right) dt = -\frac{5}{t} + \frac{4}{3} t^3 + C$$ where \(C = C_1 + C_2\).

Step 3: Check our work by differentiating

Now, we will differentiate our result to see if it matches the original function. $$\frac{d}{dt}\left(-\frac{5}{t} + \frac{4}{3} t^3 + C\right) = \frac{5}{t^2} + 4t^2$$ Since our differentiation matches the original function, our indefinite integral is correct. Our final answer is: $$\int\left(\frac{5}{t^{2}}+4 t^{2}\right) dt = -\frac{5}{t} + \frac{4}{3} t^3 + C$$