If the value of the integral int_{1}^{2} e^{x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int_{e}^{e^4} \sqrt{\ln x} dx$. Substitute $\ln x = t^2$,which implies $x = e^{t^2}$. Then $dx = 2t e^{t^2} dt$. When $x = e$,$t^2 = \ln

Vedclass · Accepted Answer

$2e^4 - e - \alpha$

Vedclass · Answer

(B) Let $I = \int_{e}^{e^4} \sqrt{\ln x} dx$. Substitute $\ln x = t^2$,which implies $x = e^{t^2}$. Then $dx = 2t e^{t^2} dt$. When $x = e$,$t^2 = \ln e = 1$,so $t = 1$. When $x = e^4$,$t^2 = \ln e^4 = 4$,so $t = 2$. Thus,$I = \int_{1}^{2} t (2t e^{t^2}) dt = \int_{1}^{2} 2t^2 e^{t^2} dt$. Using integration by parts,let $u = t$ and $dv = 2t e^{t^2} dt$. Then $du = dt$ and $v = e^{t^2}$. Applying the formula $\int u dv = uv - \int v du$: $I = [t e^{t^2}]_{1}^{2} - \int_{1}^{2} e^{t^2} dt$. $I = (2 e^{2^2} - 1 e^{1^2}) - \alpha$. $I = 2e^4 - e - \alpha$.

If the value of the integral $\int_{1}^{2} e^{x^2} dx$ is $\alpha$,then the value of $\int_{e}^{e^4} \sqrt{\ln x} dx$ is:

Similar Questions

If $I_n = \int \sin^n x \, dx$, then $n I_n - (n-1) I_{n-2}$ equals

If $\int \frac{3 e^x-7 e^{-x}}{7 e^x+3 e^{-x}} d x=K x+L \log \left(e^{-2 x}+\frac{7}{3}\right)+C$,then $K+L=$

If $\int \frac{5 \tan x}{\tan x-2} dx=ax+b \log |\sin x-2 \cos x|+c$,then $a+b=$

$\int \frac{dx}{(1 + x^2)\sqrt{1 - x^2}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module