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\(\int_0^{\pi / 2}|\sin t-\cos t| d t=\)
Options:
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2215 Upvotes
Verified Answer
The correct answer is:
\(2(\sqrt{2}-1)\)
\(\begin{aligned}
I= & \int_0^{\pi / 2}|\sin t-\cos t| d t=2 \int_0^{\pi / 4}(\cos t-\sin t) d x \\
& \left\{\because \int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x \text { if } f(2 a-x)=f(x)\right\} \\
= & 2[\sin t+\cos t]_0^{\pi / 4} \\
= & 2\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-1\right]=2(\sqrt{2}-1)
\end{aligned}\)
Hence, option (b) is correct.
I= & \int_0^{\pi / 2}|\sin t-\cos t| d t=2 \int_0^{\pi / 4}(\cos t-\sin t) d x \\
& \left\{\because \int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x \text { if } f(2 a-x)=f(x)\right\} \\
= & 2[\sin t+\cos t]_0^{\pi / 4} \\
= & 2\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-1\right]=2(\sqrt{2}-1)
\end{aligned}\)
Hence, option (b) is correct.
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