Question

Area bounded by the curve $y = xe^{x^{2}},$ $x -$axis and the ordinates $x = 0,x = a$

• A. $\frac{e^{a^{2}} + 1}{2}$sq. unit
• B. $\frac{e^{a^{2}} - 1}{2}$sq. unit
• C. $e^{a^{2}} + 1$sq. unit
• D. $e^{a^{2}} - 1$sq. unit

Answer 1

Answer: (B)

$\frac{e^{a^{2}} - 1}{2}$sq. unit

Answer 2

Required area is $\int_{0}^{a}{ydx = \int_{0}^{a}{xe^{x^{2}}dx}}$

We put $x^{2} = t \Rightarrow dx = \frac{dt}{2x}$ as $x = 0 \Rightarrow t = 0$ and

$x = a \Rightarrow t = a^{2}$, then it reduces to

$\frac{1}{2}\int_{0}^{a^{2}}{e^{t}dt = \frac{1}{2}\lbrack e^{t}\rbrack_{0}^{a^{2}} = \frac{e^{a^{2}} - 1}{2}}$ sq. unit.