Q.Evaluate given definite integral as limit of sum: $\int_0^4 (x + e^{2x})\,dx$. OR Integrate the rational function: $\dfrac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}$.

The definite integral is evaluated using the limit-of-a-sum definition (Riemann sum with $h\to0$); the OR alternative uses polynomial division followed by partial fractions in $x^2$.

Part 1 — Evaluate $\displaystyle\int_0^4(x+e^{2x})\,dx$ as a limit of a sum

By definition, $\displaystyle\int_a^b f(x)\,dx=\lim_{h\to0}h\sum_{r=0}^{n-1}f(a+rh)$, where $h=\dfrac{b-a}{n}$, $nh=b-a$.

Here $a=0,\ b=4,\ f(x)=x+e^{2x}$, $nh=4$.

$\displaystyle\sum_{r=0}^{n-1}f(rh)=\sum_{r=0}^{n-1}\big(rh+e^{2rh}\big)=h\sum_{r=0}^{n-1}r + \sum_{r=0}^{n-1}e^{2rh}$

$\displaystyle\sum_{r=0}^{n-1}r=\dfrac{n(n-1)}{2}$, and $\displaystyle\sum_{r=0}^{n-1}e^{2rh}$ is a geometric series with ratio $e^{2h}$: $=\dfrac{e^{2nh}-1}{e^{2h}-1}=\dfrac{e^8-1}{e^{2h}-1}$.

So the sum is $S = h\left[\dfrac{h\,n(n-1)}{2} + \dfrac{e^8-1}{e^{2h}-1}\right]$.

As $h\to0$ (with $nh=4$ fixed): $h^2n(n-1)=(nh)(nh-h)\to4\cdot4=16$, so the first part $\to \dfrac{16}{2}=8$.

For the second part, $e^{2h}-1\to 2h$ as $h\to0$, so $\dfrac{h}{e^{2h}-1}\to\dfrac12$, giving $\dfrac{e^8-1}{2}$.

$\displaystyle\int_0^4(x+e^{2x})\,dx = 8+\dfrac{e^8-1}{2}=\dfrac{16+e^8-1}{2}=\dfrac{e^8+15}{2}$

(Check by direct integration: $\left[\dfrac{x^2}{2}+\dfrac{e^{2x}}{2}\right]_0^4 = 8+\dfrac{e^8}{2}-\dfrac12=\dfrac{e^8+15}{2}$ — matches.)

OR — Part 2: Integrate $\dfrac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}$

Numerator $=(x^2+1)(x^2+2)=x^4+3x^2+2$; denominator $=(x^2+3)(x^2+4)=x^4+7x^2+12$. Since both are degree 4:

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