evaluate.\n int_{-3}^{4} \frac{dt}{4 + t} \n int_{-3}^{4} \frac{dt}{4 + t} = square (type an exact answer.)

evaluate.\n int_{-3}^{4} \frac{dt}{4 + t} \n int_{-3}^{4} \frac{dt}{4 + t} = square (type an exact answer.)

evaluate.\n int_{-3}^{4} \frac{dt}{4 + t} \n int_{-3}^{4} \frac{dt}{4 + t} = square (type an exact answer.)

Answer

Explanation:

Step1: Use substitution

Let $u = 4 + t$, then $du=dt$. When $t=-3$, $u = 4+( - 3)=1$; when $t = 4$, $u=4 + 4=8$.

Step2: Rewrite the integral

The integral $\int_{-3}^{4}\frac{dt}{4 + t}$ becomes $\int_{1}^{8}\frac{du}{u}$.

Step3: Integrate

We know that $\int\frac{du}{u}=\ln|u|+C$. So, $\int_{1}^{8}\frac{du}{u}=\left[\ln(u)\right]_{1}^{8}$.

Step4: Evaluate the definite - integral

$\left[\ln(u)\right]_{1}^{8}=\ln(8)-\ln(1)$. Since $\ln(1) = 0$, the result is $\ln(8)$.

Answer:

$\ln(8)$