Question

fundamental theorem of calculus: problem 3
(1 point)
evaluate the integral
integral =
\\( \int _ { 0 } ^ { \pi } ( - 3 \sin ( t ) + 8 \cos ( t ) ) d t \\)

Explanation:

Step1: Find the antiderivative

The antiderivative of \(-3\sin(t)\) is \(3\cos(t)\) (since \(\int\sin(t)dt = -\cos(t)+C\)), and the antiderivative of \(8\cos(t)\) is \(8\sin(t)\) (since \(\int\cos(t)dt=\sin(t)+C\)). So the antiderivative of \(-3\sin(t)+8\cos(t)\) is \(F(t)=3\cos(t)+8\sin(t)\).

Step2: Apply the Fundamental Theorem of Calculus

By the Fundamental Theorem of Calculus \(\int_{a}^{b}f(t)dt=F(b)-F(a)\). Here \(a = 0\), \(b=\pi\).

$$
LATEXBLOCK0
$$

Answer:

\(-6\)