Question

Find the area of the region $\rm \left\{(x, y) \mid 0 \leq y \leq x^{2}+1,0 \leq y \leq x+1,0 \leq x \leq 2\right\}$
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Answer 1

$\large\underline{\sf{Solution-}}$

The given region is

$\rm \: \left\{(x, y) \mid 0 \leq y \leq x^{2}+1,0 \leq y \leq x+1,0 \leq x \leq 2\right\}$

So, given curves are

$\rm \: y = {x}^{2} + 1 - - - (1)$

and

$\rm \: y = x + 1 - - - (2)$

Step :- 1 Point of intersection of two curves

$\rm \: {x}^{2} + 1 = x + 1$

$\rm \: {x}^{2} = x$

$\rm \: {x}^{2} - x = 0$

$\rm \: x(x - 1) = 0$

$\bf\implies \:x = 0 \: \: or \: \: x = 1$

Hᴇɴᴄᴇ,

➢ Pair of points of intersection of two curves are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 1 \\ \\ \sf 1 & \sf 2 \end{array}} \\ \end{gathered}$

Step :- 2 Curve Sketching

Curve y = $\rm \: x^2+1$ represents upper parabola having vertex at (1, 0) and axis along y - axis.

Curve y = x + 1 is a straight line which passes through the point (0, 1) and (1, 2) respectively.

{ See the attachment }

Step :- 3 Required Area

Required area bounded by the region

$\rm \: \left\{(x, y) \mid 0 \leq y \leq x^{2}+1,0 \leq y \leq x+1,0 \leq x \leq 2\right\}$

$\rm \: = \displaystyle\int_{0}^{1}\rm y_{(parabola)}dx \: + \: \displaystyle\int_{1}^{2}\rm y_{(line)}dx$

$\rm \: = \displaystyle\int_{0}^{1}\rm ( {x}^{2} + 1) \: dx \: + \: \displaystyle\int_{1}^{2}\rm (x + 1) \: dx$

$\rm \: = \bigg[\dfrac{ {x}^{3} }{3} + x \bigg]_{0}^{1} + \bigg[\dfrac{ {x}^{2} }{2} + x\bigg]_{1}^{2}$

$\rm \: = \bigg[\dfrac{1}{3} + 1 - 0 - 0 \bigg] + \bigg[\dfrac{4}{2} + 2 - \frac{1}{2} - 1\bigg]$

$\rm \: = \bigg[\dfrac{4}{3} \bigg] + \bigg[2 + 1 - \frac{1}{2}\bigg]$

$\rm \: = \dfrac{4}{3} + 3 - \frac{1}{2}$

$\rm \: = \dfrac{8 + 18 - 3}{6}$

$\bf \: = \dfrac{23}{6} \: square \: units$

Answer 2

Answer:

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