If I_1 = int_0^{pi / 4} sin^2 x , dx and I_2 | vedclass

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(B) Given: $I_1 = \int_0^{\pi / 4} \sin^2 x \, dx$ and $I_2 = \int_0^{\pi / 4} \cos^2 x \, dx$. In the interval $x \in (0, \pi / 4)$, we know that $\s

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$I_1 < I_2$

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(B) Given: $I_1 = \int_0^{\pi / 4} \sin^2 x \, dx$ and $I_2 = \int_0^{\pi / 4} \cos^2 x \, dx$. In the interval $x \in (0, \pi / 4)$, we know that $\sin x Since both $\sin x$ and $\cos x$ are positive in this interval, squaring both sides preserves the inequality: $\sin^2 x Integrating both sides over the interval $[0, \pi / 4]$: $\int_0^{\pi / 4} \sin^2 x \, dx Therefore, $I_1 < I_2$.

If $I_1 = \int_0^{\pi / 4} \sin^2 x \, dx$ and $I_2 = \int_0^{\pi / 4} \cos^2 x \, dx$, then,

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