\( y=1+|x+1| = \begin{cases}1+(x+1), x \geq-1 \\ 1-(x+1), \quad x<-1\end{cases} \)
\( =\left\{\begin{array}{cc} 2+x, & x \geq-1 \\ -x, & x<-1 \end{array}\right. \)
Required Area = Area of shaded region FCDE + Area of shaded region ABCD
\( =\int_{-3}^{-1} y_{1} d x+\int_{-1}^{3} y_{2} d x \)
\(=\int_{-3}^{-1}-x d x+\int_{-1}^{3}(x+2) d x \)
\(=\left[-\frac{x^{2}}{2}\right]_{-3}^{-1}+\left[\frac{x^{2}}{2}+2 x\right]_{-1}^{3} \)
\(=-\left[\frac{1}{2}-\frac{9}{2}\right]+\left[\frac{9}{2}+6-\frac{1}{2}+2\right] \)
\(=-\left(-\frac{8}{2}\right)+\frac{8}{2}+8 \)
= 4 + 4 + 8 = 16 sq. units
Hence required area = 16 sq. units

The given curve |x| + |v| = 1

In I quadrant, x>0, y>0
Then the curve is x + y = 1 which represent a line.
In ll quadrant x<0, y>0
Then the curve is - x + y = 1 which represents a line.
In lll quadrant x<0, y<0
Then the curve is -x - y = 1 which represents a line.
In lV quadrant x>0, y<0
Then the curve is x - y = 1 which represents a line.
When we plot above lines, then they form a square say ABCD.
Required Area = 4 (Area shaded region in the I quadrant)
\( =4 \int_{0}^{1} y(\text { line } A B) d x \)
\( =4 \int_{0}^{1}(1-x) d x \)
\(=4\left[x-\frac{x^{2}}{2}\right]_{0}^{1} \)
\(=4\left[1-\frac{1}{2}\right]=4 \times \frac{1}{2}=2 \)
Hence the required Are a = 2 sq. units

Given curves are y = x² and 
\(\mathrm{y}=|\mathrm{x}|= \begin{cases}x & \text { if } x \geq 0 \\ -x & \text { if } x<0\end{cases}\)
In I quadrant, equation of the curye is y = x and in II quadrant, equation of the curve is y = - x. The points of intersection of the parabola and curves are (-1, 1) and (1, 1). As the area to be found in symmetrical about y axis.
Required Area = 2 (Area of shaded region in the l^st quadrant)

\( =2 \int_{0}^{1}[y(\text { line })-y(\text { parabola })] d x \)
\( =2 \int_{0}^{1}\left(x-x^{2}\right) d x \)
\( =2\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1} \)
\(=2\left[\frac{1}{2}-\frac{1}{3}-0\right] \)
\(=2 \frac{1}{6}=\frac{1}{3} \)
Required Area = \(\frac{1}{3}\)units

Given equation of ellipse is 9x² + y² = 36
Which can be expressed as \(\frac{x^{2}}{4}+\frac{y^{2}}{36}=1\)
 \( \frac{y^{2}}{36} =1-\frac{x^{2}}{4} \)
\( =\frac{4-x^{2}}{4} \)
\(y^{2} =\frac{36}{4}\left(4-x^{2}\right) \)
\(y= 3 \sqrt{4-x^{2}} \).........(1)
Considering y in I quadrant

Now equation of chord AB is 
\( y-0 =\frac{6-0}{0-2}(x-2) \)
y = -3x + 6
Required Area  \( =\int_{0}^{2}[y \text { ellipse }-y \text { chord }] d x \)
\( =\int_{0}^{2}\left[3 \sqrt{4-x^{2}}-(-3 x+6)\right] d x \)
\( =3 \int_{0}^{2} \sqrt{4-x^{2}} d x-\int_{0}^{2}(6-3 x) d x \)
\( =3\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{0}^{4}-\left[6 x-\frac{3 x^{2}}{2}\right]_{0}^{2} \)
\( =3\left[0+2 \sin ^{-1}(1)-0\right]-[12-3 \times 2-0] \)
\( =6 \times \frac{\pi}{2}-6=3 \pi-6 \) sq. units

Find the intersecting point of the curyes x² + y² = 32 and y = x

\( x^{2}+y^{2}=32 \)
\(x^{2}+x^{2}=32 \)
\(2 x^{2}=32 \)
\(x^{2}=16 \)
Intersecting point is (4, 4)
Radius of the circle x² + y² = 32 is \(4 \sqrt{2}\)
Area required
\( =\int_{0}^{4} y_{1} d x+\int_{4}^{4 \sqrt{2}} y_{2} d x \)
\( =\int_{0}^{4} x d x+\int_{4}^{4 \sqrt{2}} \sqrt{32-x^{2}} d x \)
\( =\left[\frac{x^{2}}{2}\right]_{0}^{4}+\left[\frac{x}{2} \sqrt{32-x^{2}}+\frac{32}{2} \sin -1 \frac{x}{4 \sqrt{2}}\right]_{4}^{4 \sqrt{2}} \)
\(=\frac{16}{2}+\left[\frac{32}{2} \times \frac{\pi}{2}-\frac{4}{2} \sqrt{16}-\frac{32}{2} \sin \frac{4}{4 \sqrt{2}}\right] \)
\( =8+\left[8 \pi-8-\frac{16 \pi}{4}\right] \)
\(=8+4 \pi-8 \)
Required Area = 4\(\pi\) sq. units

Given equation of ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\)
\(a^{2}=9, b^{2}=4\)

\( \frac{y^{2}}{4} =1-\frac{x^{2}}{9} \)
\(\frac{y^{2}}{4} =\frac{9-x^{2}}{9} \)
\(y^{2} =\frac{4}{9}\left(9-x^{2}\right) \)
\(y =\frac{2}{3} \sqrt{9-x^{2}} \)
Area required  \(=\int_{a}^{h} y d x\)
\( =4 \int_{0}^{3} \frac{2}{3} \sqrt{9-x^{2}} d x \)
\( =\frac{8}{3} \int_{0}^{3} \sqrt{9-x^{2}} d x \)
\( =\frac{8}{3}\left[\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1} \frac{x}{3}\right]_{0}^{3} \)
\(=\frac{8}{3}\left(0+\frac{9}{2} \times \frac{\pi}{2}-0\right)=\frac{8}{3} \times \frac{9 \pi}{4}\)
Area required = 6\(\pi\) sq.unit

\(12\sqrt { 3 } \)

πa²

3πa²

prove.

\(\frac { ab\left( \pi -2 \right) }{ 4 } \)

\(\frac { 2\pi }{ 3 } -\frac { \sqrt { 3 } }{ 2 } \)

\(\frac { { 4a }^{ 2 } }{ 3 } sq.units\)

e+e^-1-2

\(\frac { 2 }{ e } \)