Mathematics

$$\displaystyle \int_0^{\pi} sin x =\displaystyle \lim_{n \rightarrow \infty} \displaystyle \Sigma_{i=1}^n sin \left(\dfrac{\pi i}{n}\right) \dfrac{\pi}{n}$$
State whether the above statement is True or False?


ANSWER

True


SOLUTION
We have to state whether $$\int_{0}^{\pi} sin x dx$$ can be represented as $$\lim_{n \rightarrow \infty} \sum_{i=1}^{n} sin \left(\dfrac{\pi i}{n}\right)\dfrac{\pi}{n} $$ is true or false.

We know that $$\int_{a}^{b} f(x) dx=\lim_{n \rightarrow \infty} \left[\dfrac{b-a}{n} \sum_{i=1}^{n} f\left(a+i\dfrac{b-a}{n}\right)\right]$$

Here we have $$f(x)=sin x,a=0,b=\pi$$

Therefore $$\int_{0}^{\pi} sin x dx=\lim_{n \rightarrow \infty} \left[\dfrac{\pi-0}{n} \sum_{i=1}^{n} f\left(0+i\dfrac{\pi-0}{n}\right)\right]$$

                                 $$=\lim_{n \rightarrow \infty} \left[\dfrac{\pi}{n} \sum_{i=1}^{n} f\left(\dfrac{\pi i}{n}\right)\right]$$

                                 $$=\lim_{n \rightarrow \infty} \left[\dfrac{\pi}{n} \sum_{i=1}^{n} sin \left(\dfrac{\pi i}{n}\right)\right]$$

            $$\int_{0}^{\pi} sin x dx=\lim_{n \rightarrow \infty} \sum_{i=1}^{n} sin \left(\dfrac{\pi i}{n}\right)\dfrac{\pi}{n} $$

Hence the answer is TRUE.
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TRUE/FALSE Hard Published on 17th 09, 2020
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