JEE Advanced · Mathematics · 27. Definite Integration
Paragraph:
Read the following passage and answer the questions.
For every function \(f(x)\) which is twice differentiable, these will be good approximation of \(\int_a^b f(x) d x \cong\left(\frac{b-a}{2}\right)\{f(a)+f(b)\}\). Now, if we take \(c=\frac{a+b}{2}\), then using above again, we get \(\int_a^b f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x \cong \frac{b-a}{4}\{f(a)+f(b)+2 f(c)\}\) and so on.
We get approximation for value of \(\int_a^b f(x) d x\).Question:
Good approximation of \(\int_0^{\pi / 2} \sin x d x\), is
- A
\(\frac{\pi}{4}\)
- B
\(\frac{\pi}{4}(\sqrt{2}+1)\)
- C
\(\frac{\pi}{8}(\sqrt{2}+1)\)
- D
\(\frac{\pi}{8}\)
Answer & Solution
Correct Answer
(C)
\(\frac{\pi}{8}(\sqrt{2}+1)\)
Step-by-step Solution
Detailed explanation
\(\int_0^{\pi / 2} \sin x d x=\frac{\frac{\pi}{2}-0}{4}\left(\sin 0+\sin \left(\frac{\pi}{2}\right)+2 \sin \left(\frac{0+\frac{\pi}{2}}{2}\right)\right)=\frac{\pi}{8}(1+\sqrt{2})\).
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