Question

question 13 · 1 point evaluate the definite integral given below. $int_{0}^{\frac{pi}{3}}(-3sin(x)-cos(x))dx$ enter an exact answer. provide your answer below:

Explanation:

Step1: Find antiderivative

The antiderivative of \(- 3\sin(x)\) is \(3\cos(x)\) and the antiderivative of \(-\cos(x)\) is \(-\sin(x)\). So the antiderivative of \(-3\sin(x)-\cos(x)\) is \(F(x)=3\cos(x)-\sin(x)\).

Step2: Apply fundamental theorem

By the fundamental theorem of calculus \(\int_{a}^{b}f(x)dx = F(b)-F(a)\), where \(a = 0\), \(b=\frac{\pi}{3}\), \(F(x)=3\cos(x)-\sin(x)\). Then \(F(\frac{\pi}{3})=3\cos(\frac{\pi}{3})-\sin(\frac{\pi}{3})=3\times\frac{1}{2}-\frac{\sqrt{3}}{2}=\frac{3 - \sqrt{3}}{2}\), and \(F(0)=3\cos(0)-\sin(0)=3\times1 - 0=3\).

Step3: Calculate the result

\(\int_{0}^{\frac{\pi}{3}}(-3\sin(x)-\cos(x))dx=F(\frac{\pi}{3})-F(0)=\frac{3 - \sqrt{3}}{2}-3=\frac{3 - \sqrt{3}-6}{2}=\frac{- 3-\sqrt{3}}{2}\).

Answer:

\(\frac{-3 - \sqrt{3}}{2}\)