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Mathematics
Mathematics
Problem
Without evaluating either integral, explain why
∫
0
π
/
2
sin
x
d
x
>
∫
0
π
/
2
2
x
π
d
x
.
\int_{0}^{\pi/2} \sin x\,dx > \int_{0}^{\pi/2} \frac{2x}{\pi}\,dx.
∫
0
π
/2
sin
x
d
x
>
∫
0
π
/2
π
2
x
d
x
.
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