Question

If the value of the integral$\int_{1}^{2}e^{x^{2}}$dx = a, then the value of $\int_{e}^{e^{4}}\sqrt{\mathcal{l}nx}$ dx is

• A. e⁴ – e – a
• B. 2e⁴ – e – a
• C. 2(e⁴ – e) – a
• D. None

Answer 1

Answer: (B)

2e⁴ – e – a

Answer 2

Let lnx = t²

x = $e^{t^{2}}$ Ž dx = $e^{t^{2}}$. 2t dt

$2\int_{1}^{2}{t^{2}.e^{t^{2}}.dt}$ = $\int_{1}^{2}{t(2te^{t^{2}})dt}$

= $(te^{t^{2}})_{1}^{2}$ – $\int_{1}^{2}{e^{t^{2}}dt}$ = 2e⁴ – e – a