Question

solve this quickly
...​

Answer 1

Answer:

Hope it helps you.....

Answer 2

Step-by-step explanation:

Given:

$\int\limits^3_1 {(x^2 + 3x + e^x)} \, dx$

$Let \ I = \int\limits^3_1 {(x^2 + 3x + e^x)} \, dx$

and

f(x) = x² + 3x + eˣ

On comparing, we get

a = 1, b = 3.

∴ h = (b - a)/n = 2/n.

$I = \int\limits^3_1 {(x^2 + 3x + e^x)} \, dx$

$\Rightarrow \lim_{h \to 0} h[f(1) + f(1 + h) + f(1 + 2h) + .. +f(1 + (n - 1) * h]$

$\Rightarrow \lim_{h \to 0}[4 + e + (1 + h)^2 + .. (1 + (n - 1)h)^2 + 3(1 + (n - 1) - h)$

$\Rightarrow \lim_{h \to 0} h[4n + e + h^2(1^2 + 2^2 + 3^2 + .. +(n - 1)^2] + 2h[1 + 2 + 3+..+(n - 1) + 3h(1 + 2 + 3... + (n - 1) + e * (e^h + ... e^{(n - 1) * h)}]$

$\Rightarrow \lim_{h \to 0} h[4n + e + h^2 \frac{(n-1)(n-2)(n-3)}{6} + 2h\frac{n(n-1)}{2}+ 3h\frac{n(n-1)}{2} + e e^h\frac{[e^{h(n-1)} - 1}{e^h - 1}]$

$\Rightarrow 8 + 8/3 + 4 + 6 + e^3 - e$

$\Rightarrow \boxed{\frac{62}{3} + e^3 - e}}$

Hope it helps!