Let \(u={ x }^{ 3 }\quad dv={ e }^{ -2x }\)
\(u_1=3{ x }^{ 2 }{ v }_{ 1 }=\frac { { e }^{ -2x } }{ -2 } \)
\(u_2=6x\quad { v }_{ 2 }=\frac { { e }^{ -2x } }{ 4 } \)
\(u_3=6\quad { v }_{ 3 }=\frac { { e }^{ -2x } }{ -8 } \)
\(u_4 = 0 \quad \ { v }_{ 4 }=\frac { { e }^{ -2x } }{ 16 } \)
Bernoulli's' formula:
\(\int { uvdx } ={ uv }_{ 1 }-u'{ v }_{ 2 }+u"{ v }_{ 3 }-u"'{ v }_{ 4 }+...\)
\(\therefore \int _{ 0 }^{ 1 }{ { x }^{ 3 }{ e }^{ -2x }dx={ \left[ { x }^{ 3 }\left( \frac { { e }^{ -2x } }{ -2 } \right) -3{ x }^{ 2 }\left( \frac { { e }^{ -2x } }{ 4 } \right) +6x\left( \frac { { e }^{ -2x } }{ -8 } \right) -6\left( \frac { { e }^{ -2x } }{ 16 } \right) \right] }_{ 0 }^{ 1 } } \)
\(={ \left[ { e }^{ -2x }\left( \frac { -{ x }^{ 3 } }{ 2 } -\frac { -3{ x }^{ 2 } }{ 4 } -\frac { 3x }{ 4 } -\frac { 3 }{ 8 } \right) \right] }_{ 0 }^{ 1 }\)
\(=\left[ { e }^{ -2 }\left( -\frac { 1 }{ 2 } -\frac { 3 }{ 4 } -\frac { 3 }{ 4 } -\frac { 3 }{ 8 } \right) -{ e }^{ -0 }\left( \frac { -3 }{ 8 } \right) \right] \)
\(={ e }^{ -2 }\left( \frac { -4-12-3 }{ 8 } \right) +\frac { 3 }{ 8 } \)
\(={ e }^{ -2 }\left( \frac { -19 }{ 8 } \right) +\frac { 3 }{ 8 } =\frac { 3 }{ 8 } -\frac { 19 }{ 8 } { e }^{ -2 }\)

Volume of the right circular conical frustum is obtained by revolving the line y = x between x = a and x = b around the x - axis

\(\therefore\) Height of the frustum h = b - a
\(\therefore\)Volume \(=\pi \int _{ a }^{ b }{ { x }^{ 2 }dx } =\pi { \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ a }^{ b }\)
\(=\frac { \pi }{ 3 } [{ b }^{ 3 }-{ a }^{ 3 }]\)
\(=\frac { \pi }{ 3 } (b-a)({ b }^{ 2 }+ab+{ a }^{ 2 })\)
Now, substitute h = b - a, r = a and R = b we get Volume of the conical frustum
\(\frac { \pi }{ 3 } [h({ R }^{ 2 }+rR+{ r }^{ 2 })]\)
Given h = 2 m, r = 1 m, R = 2 m we get
Required volume \(=\frac { \pi }{ 3 } [2(4+2+1)]\)
\(=\frac { \pi }{ 3 } (14)\)
\(=\frac { 14\pi }{ 3 } \)

Equation of the given curves are y² = 4x and x² = 4y

\(\therefore \) Required area \(=\int _{ 0 }^{ 4 }{ \left( \sqrt { 4x } -\frac { { x }^{ 2 } }{ 4 } \right) dx } \)
\(=\int _{ 0 }^{ 4 }{ \left( 2\sqrt { x } -\frac { { x }^{ 2 } }{ 4 } \right) dx } \)
\(={ \left[ \frac { { 2x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ 4 }{ \left[ \frac { 4 }{ 3 } x\sqrt { x } -\frac { { x }^{ 3 } }{ 12 } \right] }_{ 0 }^{ 4 }\)
\(=\frac { 4 }{ 3 } (4)(2)-\frac { 64 }{ 12 } \)
\(=\frac { 32 }{ 3 } -\frac { 32 }{ 6 } =\frac { 64-32 }{ 6 } =\frac { 32 }{ 6 } \)
\(=\frac { 16 }{ 3 } \) sq.units
Yes the area can be divided into 3 equal parts and the area to the divided among his, wife daughter and son is  \(=\frac { 16 }{ 3 } \)sq.units

The lines 5x − 2y = 15, x + y + 4 = 0 intersect at (1, −5). The line 5x − 2y =15 meets the x-axis at (3, 0). The line x + y + 4 = 0 meets the x-axis at (−4, 0). The required area is shaded. It lies below the x-axis. It can be computed either by considering vertical strips or horizontal strips.
When we do by vertical strips, the region has to be divided into two sub-regions by the line x = 1. Then, we get

\(A=\left| \int _{ -4 }^{ 1 }{ ydx } \right| +\left| \int _{ 1 }^{ 3 }{ ydx } \right| \)
\(=\left| \int _{ -4 }^{ 1 }{ (-4-x)dx } \right| +\left| \int _{ 1 }^{ 3 }{ \left( \frac { 5x-15 }{ 2 } \right) dx } \right| \)
\(=\left| { \left( -4x-\frac { { x }^{ 2 } }{ 2 } \right) }_{ -4 }^{ 1 } \right| +\left| { \left( \frac { { 5x }^{ 2 } }{ 4 } -\frac { 15x }{ 2 } \right) }_{ 1 }^{ 3 } \right| \)
\(=\left| \left( -\frac { 9 }{ 2 } \right) -\left( 8 \right) \right| +\left| \left( -\frac { 45 }{ 4 } \right) -\left( -\frac { 25 }{ 4 } \right) \right| \)
\(=\frac { 25 }{ 2 } +5\)
\(=\frac { 35 }{ 2 } \)
When we do by horizontal strips, there is no need to subdivide the region. In this case, the area is bounded on the right by the line 5x − 2y = 15 and on the left by x + y + 4 = 0. So, we get
\(A=\int _{ -5 }^{ 0 }{ [{ x }_{ R }-{ x }_{ L }]dy } =\int _{ -5 }^{ 0 }{ \left[ \frac { 15+2y }{ 5 } -(-4-y) \right] dy } \)
\(=\int _{ -5 }^{ 0 }{ \left[ 7+\frac { 7y }{ 5 } \right] dy={ \left[ 7y+\frac { { 7y }^{ 2 } }{ 10 } \right] }_{ -5 }^{ 0 } } \)
\(\\ =0-\left[ -35+\frac { 35 }{ 2 } \right] =\frac { 35 }{ 2 } \)

The smaller segment is sketched. Here 0

\(A=2\int _{ h }^{ a }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } dx=2{ \left[ \frac { x\sqrt { { a }^{ 2 }-{ x }^{ 2 } } }{ 2 } +\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }\left( \frac { x }{ a } \right) \right] }_{ h }^{ a }\)
\(=2\left[ 0+\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }(1) \right] -2\left[ \frac { h\sqrt { { a }^{ 2 }-{ h }^{ 2 } } }{ 2 } +\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }\left( \frac { h }{ a } \right) \right] \)
\(={ a }^{ 2 }\left( \frac { \pi }{ 2 } \right) -h\sqrt { { a }^{ 2 }-{ h }^{ 2 } } -{ a }^{ 2 }{ sin }^{ -1 }\left( \frac { h }{ a } \right) \)
\(=a^{2}\left[\frac{\pi}{2}-\sin ^{-1}\left(\frac{h}{a}\right)\right]-h \sqrt{a^{2}-h^{2}}\)
\(={ a }^{ 2 }{ cos }^{ -1 }\left( \frac { h }{ a } \right) -h\sqrt { { a }^{ 2 }-{ h }^{ 2 } } \)