Question

Let c > 0 be such that
$\int_{0}^{c} e^{s^{2}} d s = 3$
Then, the value of
$\int_{0}^{c} (\int_{x}^{c} e^{x^{2} + y^{2}} d y) d x$
equals _____________ (rounded off to two decimal places).

Answer: $\underset{_____________}{4.50}$

Solution :-

Let us consider $I = \int_{0}^{c} \int_{0}^{c} e^{x^{2} + y^{2}} d x d y \dots \dots (1)$

Here, put $\begin{array}{llllllllll} x & = & r cos θ \\ y & = & r sin θ \end{array}$

\begin{array}{llllllllll} x^{2} + y^{2} & = & {(r cos θ)}^{2} + {(r sin θ)}^{2} \\ = & r^{2} {cos}^{2} θ + r^{2} {sin}^{2} θ \\ = & r^{2} ({cos}^{2} θ + {sin}^{2} θ) \\ = & r^{2} . \end{array}

Now,

\begin{array}{llllllllll} J \frac{(x, y)}{(r, θ)} = | \begin{array}{cccccccccc} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial θ} & \frac{\partial y}{\partial θ} \end{array} | & = & | \begin{array}{cccccccccc} cos θ & sin θ \\ - r sin θ & r cos θ \end{array} | \\ = & r {cos}^{2} θ - (- r {sin}^{2} θ) \\ = & r ({cos}^{2} θ + {sin}^{2} θ) \\ = & r . \end{array}

In eq(1), because of inner and outer integration limit goes to $0 \to c$ here $c > 0$ (given) hence, integration limit for $r$ goes to $0 \to c$ and integration limit for $θ$ goes to $0 \to \frac{π}{2} .$

Now, apply change of variable in eq(1), we get

\begin{array}{llllllllll} I & = & \int_{0}^{\frac{π}{2}} \int_{0}^{c} e^{r^{2}} ∣ J ∣ d r d θ \\ I & = & \int_{0}^{\frac{π}{2}} \int_{0}^{c} e^{r^{2}} r d r d θ \\ = & \int_{0}^{\frac{π}{2}} d θ \cdot \frac{1}{2} \int_{0}^{c} e^{r^{2}} 2 r d r \\ = & {[θ]}_{0}^{\frac{π}{2}} \cdot \frac{1}{2} {[e^{r^{2}}]}_{0}^{c} = \frac{π}{4} (e^{c^{2}} - 1) . \end{array}

∴ \int_{0}^{c} \int_{0}^{c} e^{x^{2} + y^{2}} d x d y = \frac{π}{4} (e^{c^{2}} - 1)

Now, we write above as

\begin{array}{llllllllll} \int_{0}^{c} \int_{0}^{c} e^{x^{2}} \cdot e^{y^{2}} d x d y & = & \frac{π}{4} (e^{c^{2}} - 1) \\ \int_{0}^{c} e^{y^{2}} d y \cdot \int_{0}^{c} e^{x^{2}} d x & = & \frac{π}{4} (e^{c^{2}} - 1) \\ \int_{0}^{c} e^{x^{2}} d x \cdot \int_{0}^{c} e^{x^{2}} d x & = & \frac{π}{4} (e^{c^{2}} - 1) (r e p l a c e d y w i t h x) \\ {(\int_{0}^{c} e^{x^{2}} d x)}^{2} & = & \frac{π}{4} (e^{c^{2}} - 1) \\ ∴ \int_{0}^{c} e^{x^{2}} d x & = & \frac{\sqrt{π}}{2} {(e^{c^{2}} - 1)}^{\frac{1}{2}} . \\ Now, \int_{0}^{c} e^{s^{2}} d s & = & \frac{\sqrt{π}}{2} {(e^{c^{2}} - 1)}^{\frac{1}{2}} (r e p l a c e d x w i t h s) \\ 3 & = & \frac{\sqrt{π}}{2} {(e^{c^{2}} - 1)}^{\frac{1}{2}} (∵ \int_{0}^{c} e^{s^{2}} d s = 3 g i v e n) \\ {(e^{c^{2}} - 1)}^{\frac{1}{2}} & = & \frac{6}{\sqrt{π}} \\ (e^{c^{2}} - 1) & = & \frac{36}{π} \\ e^{c^{2}} & = & \frac{36 + π}{π} \\ c^{2} & = & log (\frac{36 + π}{π}) \\ c & = & {[log (\frac{36 + π}{π})]}^{\frac{1}{2}} \approx 1.588 \end{array}

Now, we want to find the value of $\int_{0}^{c} (\int_{x}^{c} e^{x^{2} + y^{2}} d y) d x .$
Here, because of inner integration limit goes to $x \to c$ and outer integration limit goes to $0 \to c$ hence, integration limit for $θ$ goes to $0 \to \frac{π}{4}$ and integration limit for $r$ goes to $0 \to c .$

\begin{array}{llllllllll} ∴ \int_{0}^{c} (\int_{x}^{c} e^{x^{2} + y^{2}} d y) d x & = & \int_{0}^{\frac{π}{4}} \int_{0}^{c} e^{r^{2}} r d r d θ \\ = & \int_{0}^{\frac{π}{4}} d θ \cdot \frac{1}{2} \int_{0}^{c} e^{r^{2}} 2 r d r \\ = & \frac{π}{4} \cdot \frac{1}{2} {[e^{r^{2}}]}_{0}^{c} \\ = & \frac{π}{8} (e^{c^{2}} - 1) \\ = & \frac{π}{8} (e^{{(1.588)}^{2}} - 1) \\ \approx & 4.496 . \end{array}

rounded off to two decimal places is 4.50.

So, the correct answer is $\to$ $\underset{_____________}{4.50}$

#Integral Calculus #IIT JAM PYQ 2024 #Mathematics #IIT JAM PYQs

Question

Let c > 0 be such that ∫0ces2d⁢s=3 Then, the value of ∫0c(∫xcex2+y2d⁢y)d⁢x equals _____________ (rounded off to two decimal places).

Solution :-

Let c > 0 be such that
$\int_{0}^{c} e^{s^{2}} d s = 3$
Then, the value of
$\int_{0}^{c} (\int_{x}^{c} e^{x^{2} + y^{2}} d y) d x$
equals _____________ (rounded off to two decimal places).