Question

theorem of calculus: problem 3

$$ int _ { 0 } ^ { pi } ( - 4 sin ( t ) + 1 cos ( t ) ) d t $$

Explanation:

Step1: Integrate term - by - term

$$\int_{0}^{\pi}(-4\sin(t)+\cos(t))dt=-4\int_{0}^{\pi}\sin(t)dt+\int_{0}^{\pi}\cos(t)dt$$
We know that $\int\sin(t)dt =-\cos(t)+C$ and $\int\cos(t)dt=\sin(t)+C$.

Step2: Apply the fundamental theorem of calculus

For $-4\int_{0}^{\pi}\sin(t)dt$, using $F(t)=4\cos(t)$ (since the antiderivative of $- 4\sin(t)$ is $4\cos(t)$), then $4\cos(t)\big|_{0}^{\pi}=4\cos(\pi)-4\cos(0)=4\times(- 1)-4\times1=-8$.
For $\int_{0}^{\pi}\cos(t)dt$, using $F(t)=\sin(t)$, then $\sin(t)\big|_{0}^{\pi}=\sin(\pi)-\sin(0)=0 - 0=0$.

Answer:

$-8$