Given displaystyle int - 1 - 2 - - e - - x - 2 - - - dx-a- the value of displaystyle int - e - - e - 4 - - sqrt - ln - left- x right- - - dx - is- e - 4 -e- e - 4 -a2- e - 4 -a2- e - 4 -e-a

Question

Given displaystyle int - 1 - 2 - - e - - x - 2 - - - dx-a- the value of displaystyle int - e - - e - 4 - - sqrt - ln - left- x right- - - dx -  is- e - 4 -e- e - 4 -a2- e - 4 -a2- e - 4 -e-a

Toppr Admin · Accepted Answer

Given -x222B-21ex2dx-a-xA0-Let I-x222B-e4e-x221A-ln-x-dxPut ln-x-t2-x21D2-1xdx-2tdt-x2234-I-x222B-21et2-2t2dt-tet2-21-x2212-x222B-21et2dt-2e4-x2212-e-x2212-a-

Given $\int_{1}^{2} e^{x^{2}} d x = a,$ the value of $\int_{e}^{e^{4}} \sqrt{ln (x)} d x$ is?
$e^{4} - e$
$e^{4} - a$
$2 e^{4} - a$
$2 e^{4} - e - a$

Given $\int_{1}^{2} e^{x^{2}} d x = a,$
Let $I = \int_{e}^{e^{4}} \sqrt{ln (x)} d x$
Put $ln (x) = t^{2} \Rightarrow \frac{1}{x} d x = 2 t d t$
$∴ I = \int_{1}^{2} e^{t^{2}} . {2 t}^{2} d t = {(t e^{t^{2}})}_{1}^{2} - \int_{1}^{2} e^{t^{2}} d t = {2 e}^{4} - e - a .$

Given ∫21ex2dx=a, the value of ∫e4e√ln(x)dx is?e4−ee4−a2e4−a2e4−e−a

Given ∫21ex2dx=a, Let I=∫e4e√ln(x)dxPut ln(x)=t2⇒1xdx=2tdt∴I=∫21et2.2t2dt=(tet2)21−∫21et2dt=2e4−e−a.

Given $\int_{1}^{2} e^{x^{2}} d x = a,$ the value of $\int_{e}^{e^{4}} \sqrt{ln (x)} d x$ is?
$e^{4} - e$
$e^{4} - a$
$2 e^{4} - a$
$2 e^{4} - e - a$

Given $\int_{1}^{2} e^{x^{2}} d x = a,$
Let $I = \int_{e}^{e^{4}} \sqrt{ln (x)} d x$
Put $ln (x) = t^{2} \Rightarrow \frac{1}{x} d x = 2 t d t$
$∴ I = \int_{1}^{2} e^{t^{2}} . {2 t}^{2} d t = {(t e^{t^{2}})}_{1}^{2} - \int_{1}^{2} e^{t^{2}} d t = {2 e}^{4} - e - a .$