Question

\( 020 \quad 10.0 \) points Evaluate the definite integral \[ I=\int_{0}^{\pi / 2}(5 \cos x-\sin x) d x \] 1. \( I=5 \) 2. \( I=7 \) 3. \( I=6 \) 4. \( I=8 \) 5. \( I=4 \) correct Explanation: By the Fundamental Theorem of Calculus, \[ I=[F(x)]_{0}^{\pi / 2}=F\left(\frac{\pi}{2}\right)-F(0) \] for any anti-derivative \( F \) of \[ f(x)=5 \cos x-\sin x \] Taking \[ F(x)=5 \sin x+\cos x \] and using the fact that \[ \begin{array}{l} \cos 0=\sin \frac{\pi}{2}=1 \\ \sin 0=\cos \frac{\pi}{2}=0 \end{array} \]

          \( 020 \quad 10.0 \) points
Evaluate the definite integral
\[
I=\int_{0}^{\pi / 2}(5 \cos x-\sin x) d x
\]
1. \( I=5 \)
2. \( I=7 \)
3. \( I=6 \)
4. \( I=8 \)
5. \( I=4 \) correct

Explanation:
By the Fundamental Theorem of Calculus,
\[
I=[F(x)]_{0}^{\pi / 2}=F\left(\frac{\pi}{2}\right)-F(0)
\]
for any anti-derivative \( F \) of
\[
f(x)=5 \cos x-\sin x
\]

Taking
\[
F(x)=5 \sin x+\cos x
\]
and using the fact that
\[
\begin{array}{l}
\cos 0=\sin \frac{\pi}{2}=1 \\
\sin 0=\cos \frac{\pi}{2}=0
\end{array}
\]

020 10.0 points
Evaluate the definite integral

I=∫0^π / 2(5 cos x-sin x) d x

1. I=5
2. I=7
3. I=6
4. I=8
5. I=4 correct

Explanation:
By the Fundamental Theorem of Calculus,

I=[F(x)]0^π / 2=F((π)/(2))-F(0)

for any anti-derivative F of

f(x)=5 cos x-sin x

Taking

F(x)=5 sin x+cos x

and using the fact that

cos 0=sin(π)/(2)=1
sin 0=cos(π)/(2)=0

Added by Stephen C.

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