# Evaluate $\int^4_1 e^ \sqrt {x}dx$

Evaluate $\int^4_1 e^ \sqrt {x}dx$

solution:-

Here $1<x<4$

$1<\sqrt x<2$

$e<e^ \sqrt {x}<e^ 2$

$\int^4_1$e dx$<\int^4_1 e^ \sqrt {x}dx<\int^4_1 e^ 2dx$

$3e <\int^4_1 e^ \sqrt {x}dx<3 e^ 2$

But in this objective question

Options are

a)$e$

b)$e^2$

c)$2e$

d)$2e^2$

Hint: Make the substitution $x=t^2$ to get the integral $2\int_1^2 te^t \mathrm{d}t$.
HINT: $$3 e < 2e^2 < 3e^2$$ As suggested, the substitution $x=t^2$ yields $$2\int_1^2 te^t dt$$ Using integration by parts, you get $$2\int_1^2 te^t dt = 2\left([te^t]_1^2 - \int_1^2 e^t dt\right) = 2 \left( 2e^2 - e^1 -e^2 + e^1 \right) = 2 e^2$$
• So you mean ans is (d)$2e^2$ – rst Jun 3 '13 at 10:51
• I think $2e^2 > 3e$ so $3e^2>2e^2 > 3e$ – rst Jun 3 '13 at 10:55