Question

exercises 1.3 the fundamental theorem of calculus
score: 225/260 answered: 23/26
question 24
evaluate the definite integral
\\(\int_{0}^{\pi} 6\sin(x) dx\\)
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Explanation:

Step1: Find the antiderivative of \(6\sin(x)\)

The antiderivative of \(\sin(x)\) is \(-\cos(x)\), so the antiderivative of \(6\sin(x)\) is \(6\times(-\cos(x))=-6\cos(x)\).

Step2: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that \(\int_{a}^{b} f(x)dx = F(b)-F(a)\), where \(F(x)\) is the antiderivative of \(f(x)\). Here, \(a = 0\), \(b=\pi\), and \(F(x)=-6\cos(x)\). So we calculate \(F(\pi)-F(0)\):
\(F(\pi)=-6\cos(\pi)\) and \(F(0)=-6\cos(0)\).
We know that \(\cos(\pi)= - 1\) and \(\cos(0)=1\).
So \(F(\pi)-F(0)=-6\times(-1)-(-6\times1)\)
\(=6 + 6\)
\(=12\)

Answer:

\(12\)