∫_0^2 e^x dx a=0 and b=2 ∴ h = b a/n ⇒ nh = 2 and f(x) = e^x Now, ∫_0^2e^x dx = lim_h→ 0 h[f(0) + f(0+h)+f(0+2h)+...+f0+(n 1) h] ∴ I = lim_h → 0h[1+e^h+e ...

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Byju's Answer

Standard XII

Mathematics

Right Hand Limit

∫02 e x d x

Question

$\int_{0}^{2} e^{x} d x$

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Solution

$\int_{0}^{2} e^{x} d x a = 0 a n d b = 2 ∴ h = \frac{b - a}{n} \Rightarrow n h = 2 a n d f (x) = e^{x} N o w, \int_{0}^{2} e^{x} d x = {lim}_{h \to 0} h [f (0) + f (0 + h) + f (0 + 2 h) + . . . + f 0 + (n - 1) h] ∴ I = {lim}_{h \to 0} h [1 + e^{h} + e^{2 h} + . . . + e^{(n - 1) h}] = {lim}_{h \to 0} h = {lim}_{h \to 0} h [\frac{1. (e^{h})^{n} - 1}{e^{h} - 1}] = {lim}_{h \to 0} h [\frac{e^{n h} - 1}{e^{h} - 1}] = {lim}_{h \to 0} h (\frac{e^{2} - 1}{e^{h} - 1}) = e^{2} {lim}_{h \to 0} \frac{h}{e^{h} - 1} - {lim}_{h \to 0} \frac{h}{e^{h} - 1} = e^{2} - 1 = e^{2} - 1$

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Similar questions

\int_{0}^{2} e^{x} d x

Q. Evaluate

\int_{0}^{2} e^{x} d x

as the limit of a sum.

Q. Express

\int_{0}^{2} e^{x} d x

as the limit of a sum.

State whether the given statement is True or False

\int_{0}^{2} e^{x} d x

can be represented as

2 lim n \to \infty \frac{1}{n} [e^{0} + e^{\frac{2}{n}} + e^{\frac{4}{n}} + . . . . . . + e^{\frac{2 (n - 1)}{n}}]

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∫20ex dx

∫20 ex dx a=0 and b=2∴h=b−an⇒nh=2 and f(x)=exNow,∫20ex dx=limh→0 h[f(0)+f(0+h)+f(0+2h)+...+f0+(n−1) h]∴ I=limh→0h[1+eh+e2h+...+e(n−1)h]=limh→0h=limh→0h[1.(eh)n−1eh−1]=limh→0h[enh−1eh−1]=limh→0h(e2−1eh−1)=e2limh→0heh−1−limh→0heh−1=e2−1=e2−1

$\int_{0}^{2} e^{x} d x$