Mathematics-I (3110014)

Question-2c-OR

BE | Semester-1 Winter-2019 | 17-01-2020

Q2) (c)

$For, I = \int_{x = 0}^{x = 1} \int_{y = x}^{y = 1} \sin (y^{2}) dy dx$

$Old Limit (Fig . 1)$	$New Limit (Fig . 2)$
$x \to 0 to x \to 1$ $y \to x to y \to 1$	$y \to 0 to y \to 1$ $x \to 0 to x \to y$

Applying New Limits,

For, I = \int_{y = 0}^{y = 1} \int_{x = 0}^{x = y} \sin (y^{2}) dy dx

For, I = \int_{0}^{1} \sin (y^{2}) \cdot {[x]}_{0}^{y} dy

For, I = \int_{0}^{1} \sin (y^{2}) \cdot [y - 0] dy

For, I = \int_{0}^{1} y \cdot \sin (y^{2}) dy

Let, y^{2} = t

⟹ 2 y \cdot dy = dt

⟹ y \cdot dy = \frac{dt}{2}

Change in limits :

y = 0 \Rightarrow t = 0

y = 1 \Rightarrow t = 1

Substituting the values,

For, I = \int_{t = 0}^{t = 1} \sin t \frac{dt}{2}

For, I = \frac{1}{2} \int_{t = 0}^{t = 1} \sin t dt

For, I = \frac{1}{2} {[- \cos t]}_{0}^{1}

For, I = \frac{1}{2} [- \cos 1 + \cos 0]

For, I = \frac{1}{2} [1 - \cos 1]

So, I = \int_{0}^{1} \int_{x}^{1} \sin (y^{2}) dy dx = \frac{1 - \cos 1}{2}