Mathematics

Evaluate the following integral as limit of sum:
$$\displaystyle \int_{0}^{2}e^x\ dx$$


SOLUTION
We know that
$$\displaystyle \int_a^b f(x) dx=\displaystyle \lim_{h\to 0}h [f(a) +f(a+h)+...+f(a+(h-1) h)]$$, where $$h=\dfrac {b-a}{2}$$

Here $$a=0, b=2, f(x)=e^x $$ and $$h=\dfrac {2-0}{n}$$

$$\therefore \ \displaystyle \int_{0}^2 e^x \ dx =\displaystyle \lim_{h\to 0}h [f(0) +f(h) +f(2h) +...+f((n-1) )h]$$

$$=\displaystyle \lim_{h\to 0}h[e^0 +e^h +e^{2h} +....+ e^{(n-1)h}]$$

$$=\displaystyle \lim_{h\to 0} h \left [\dfrac {(e^h)^n -1}{e^h -1}\right] =\displaystyle \lim_{h\to 0} \dfrac {e^2 -1}{\left (\dfrac {e^2 -1}{h}\right)}=\dfrac {e^2 -1}{1} =e^x -1$$

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Subjective Medium Published on 17th 09, 2020
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