Evaluate the following definite integrals as limit of sums.
∫40(x+e2x)dx.
We know that ∫baf(x)dx=limh→0h[f(a)+f(a+h)+f(a+2h)+.....+f{a+(n−1)h}]
where, nh =b-a
Given,∫40(x+e2x)dx
Here, a=0, b=4 and nh =4
and f(x)=(x+e2x),∴f(0)=(0+e2(0))=1,
f(0+h)=h+e2h,f(0+2h)=2h+e4h,
and f(0+(n−1)h)=(n−1)h+e2(n−1)h
∴∫40(x+e2x)dx=limh→0h[1+(h+e2h)+(2h+e4h)+...+(n−1)h+e2(n−1)h]=limh→0h[{h+2h+3h+...+(n−1)}h+{1+e2h+e4h+...+e2(n−1)h}]
=limh→0h[h{1+2+3+...+(n−1)}+{1+e2h+e4h+...+e2nh−2h}]=limh→0h{h(n−1)n2+1((e2h)n−1e2h−1)}(∵∑n=n(n+1)2)
(∵1+e2h+e4h+...+e2(n−1)h) is a GP series whose first term is 1 and common ratio r=e2h1=e2h
∴ Sum =a(rn−1)r−1
=limh→0{h2(n−1)n2+h(e2nh−1)e2h−1}=limh→0(nh−h)(nh)2+limh→0(e2)4−1e2h−12h×2=limh→0{(4−h)42+e8−12}(∵limh→0ex−1x=1)=4×42+e8−12=(8−12)+e82=15+e82