Mathematics

Evaluate:
$$\int _{1}^{2}(x^{2}+e^{x})dx$$
using limit of sum


SOLUTION
We have, $$\displaystyle \int^{2}_{1}\left(x^{2}+e^{x}\right) \Rightarrow a=1,b=1 f\left(x\right)=x^2+e^x,h=\dfrac{1}{n}$$
$$I=\displaystyle \int^{2}_{1} \left(x^{2}+e^{x}\right)dx= \lim_{ h \rightarrow 0 } \sum_{r=1}^{h} f\left(a+rh\right)$$
$$=\displaystyle\lim_{h \rightarrow 0}h \left[f\left(1\right)+f\left(1+h\right)+f\left(1+2h\right)+....+f\left(1+\left(n-1\right)h\right)\right]$$
$$= \displaystyle \lim_{h \rightarrow 0}h\left[1+e\right]\left(\left(1+h\right)^2+e^{1+h}\right) +\left(\left(1+2h\right)^2+e^{1+h}\right) +....+\left(\left(1+(n+1)h\right)^2+e^{1+(n-1)h}\right)$$
$$ = \displaystyle \lim_{h \rightarrow 0} \left[1+e+\left(n-1\right).1 +h^2 \left(1+2^2+3^2..... (n-1)^2 \right) 2h \left(1+2+3..... (n-1) \right) \left(e^{1+h}+ e^{1+2h} +....+e^{1+(n-1)h} \right) \right]$$
$$=\displaystyle \lim_{h \rightarrow 0} h \left\{ n+e+h^2 \left(\dfrac{2\left(n-1\right)\left(n-2\right) \left(2n-3\right) } {6}\right)  +2h \dfrac{(n-1)}n{2}+ e\left[e^h . \dfrac{\left(e^{h\left(n-1\right)}-1\right)}{e^h-1}\right]\right\}$$
$$\displaystyle =\lim_{h \rightarrow 0} \left\{ n \times \dfrac{1}{n } + \dfrac{e}{n}+ \dfrac{1}{n^2}  \dfrac{\left( 2n^3...... \right)}{6} + \dfrac{2}{n^2} \left( n^2-1 \dfrac{2}{n} \dfrac{\left(n-1\right)n}{2} \right)  +e \left[ e^{\dfrac{1}{n}} \dfrac{e^\dfrac{1}{n} \left(n-1\right)-1} {e^{\dfrac{1}{n}}-1} \right]\right\}$$
$$=1+0+\dfrac{1}{3}+e\left(e^2-1\right)$$
$$=\dfrac{5}{3}+e \left(e^2-1\right)$$ Ans
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