Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
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
Solution:
1570 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{8}(\sqrt{2}+1)$
$\frac{\pi}{8}(\sqrt{2}+1)$
$\int_0^{\pi / 2} \sin x d x=\frac{\frac{\pi}{2}-0}{4}\left(\sin 0+\sin \left(\frac{\pi}{2}\right)+2 \sin \left(\frac{0+\frac{\pi}{2}}{2}\right)\right)=\frac{\pi}{8}(1+\sqrt{2})$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.