\(I=\int _{ 0 }^{ 1 }{ { x }^{ 2 } } { (1-x) }^{ 3 }dx\)
We know \(\int _{ 0 }^{ 1 }{ { x }^{ m }{ (1-x) }^{ n }dx=\frac { m!\times n! }{ (m+n+1)! } } \)
\(\therefore I=\frac { 2!\times 3! }{ (2+3+1)! } =\frac { 2\times 3\times 2\times 1 }{ 6\times 5\times 4\times 3\times 2\times 1 } \)
\(I=\frac { 1 }{ 60 } \)

\(Let\ I=\int _{ 0 }^{ \pi /2 }{ { sin }^{ 2 }x{ cos }^{ 4 }xdx } \)
\({ I }_{ m,n }=\int _{ 0 }^{ \pi /2 }{ { sin }^{ m } } x{ cos }^{ n }xdx=\frac { n-1 }{ m+n } { I }_{ m,m-2 }n\ge 2\)
\(=\left( \frac { m-1 }{ n+m } \right) \left( \frac { m-3 }{ n+m-2 } \right) \left( \frac { m-5 }{ m+m-4 } \right) ...\frac { 2 }{ n+3 } .\frac { 1 }{ n+1 } \)
Here m = 2, n = 4
\(\therefore I=\frac { 3 }{ 6 } \times \frac { 1 }{ 4 } \times \frac { 1 }{ 2 } \times \frac { \pi }{ 32 } \)

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:
\(\int { uvdx } ={ uv }_{ 1 }-u'{ v }_{ 2 }+u"{ v }_{ 3 }-u"'{ v }_{ 4 }+...\)
\(\therefore \int _{ 0 }^{ 1 }{ { x }^{ 3 }{ e }^{ -2x }dx={ \left[ { x }^{ 3 }\left( \frac { { e }^{ -2x } }{ -2 } \right) -3{ x }^{ 2 }\left( \frac { { e }^{ -2x } }{ 4 } \right) +6x\left( \frac { { e }^{ -2x } }{ -8 } \right) -6\left( \frac { { e }^{ -2x } }{ 16 } \right) \right] }_{ 0 }^{ 1 } } \)
\(={ \left[ { e }^{ -2x }\left( \frac { -{ x }^{ 3 } }{ 2 } -\frac { -3{ x }^{ 2 } }{ 4 } -\frac { 3x }{ 4 } -\frac { 3 }{ 8 } \right) \right] }_{ 0 }^{ 1 }\)
\(=\left[ { e }^{ -2 }\left( -\frac { 1 }{ 2 } -\frac { 3 }{ 4 } -\frac { 3 }{ 4 } -\frac { 3 }{ 8 } \right) -{ e }^{ -0 }\left( \frac { -3 }{ 8 } \right) \right] \)
\(={ e }^{ -2 }\left( \frac { -4-12-3 }{ 8 } \right) +\frac { 3 }{ 8 } \)
\(={ e }^{ -2 }\left( \frac { -19 }{ 8 } \right) +\frac { 3 }{ 8 } =\frac { 3 }{ 8 } -\frac { 19 }{ 8 } { e }^{ -2 }\)

Taking u = x² and v = cos nx, and applying the Bernoulli’s formula, we get
\(I=\int _{ 0 }^{ \pi }{ { x }^{ 2 }cos\quad nxdx } ={ \left[ ({ x }^{ 2 })\left( \frac { sin\quad nx }{ n } \right) -(2x)\left( -\frac { cos\ nx }{ { n }^{ 2 } } \right) +(2)\left( -\frac { sin\quad nx }{ { n }^{ 3 } } \right) \right] }_{ 0 }^{ \pi }\)
\(=\frac { 2\pi { (-1) }^{ n } }{ { n }^{ 2 } } \), since cos n\(\pi\) = (-1)ⁿ and sin n\(\pi\) = 0

\(\frac { 2 }{ a } \)

\(\frac { { 208a }^{ 2 } }{ 3 } \)

\(\frac { 13 }{ 3 } \)

\({ tan }^{ -1 }e-\frac { \pi }{ 4 } \)

\(\frac { \pi }{ 60 } \)

\(\frac { { x }^{ 3 } }{ 3 } +c\)

\(\frac { 1 }{ 2 } \left[ 1-{ e }^{ \frac { \pi }{ 2 } } \right] \)

\(\frac { 1 }{ 5 } \left( { e }^{ 5 }-2 \right) \)

Let I = \(\int _{ 0 }^{ 1 }{ \frac { { e }^{ x } }{ 1+{ e }^{ 2x } } dx } \)

Put e^x = t ⇒ e^x dx = dt
∴ \(\int _{ 1 }^{ e }{ \frac { dt }{ 1+{ t }^{ 2 } } dx } { \left[ { tan }^{ -1 }(t) \right] }_{ 1 }^{ e }\)
= tan^-1(e) - tan^-1(1)
= tan^-1(e) -\(\frac { \pi }{ 4 } \)

Given that I =\(\int ^\frac {\pi}{2}_{0} \)( sin²x + cos⁴x)dx =\(\int ^\frac {\pi}{2}_{0} \) sin²x dx+\(\int ^\frac {\pi}{2}_{0} \)cos⁴x dx\(\frac {1}{2} \times \frac {\pi}{2} + \frac {3}{4} \times \frac {1}{2} \times \frac {\pi}{2} = \frac {7\pi}{16} \)

Equation of the given curve is y + 1 = x²
\(\therefore\) The vertex of this open upward parabola is (0, -1)

Required volume \(=\int _{ -1 }^{ 3 }{ { x }^{ 2 }dy } \)
\(=\pi \int _{ -1 }^{ 3 }{ (1+y)dy } \)
\(=\pi { \left[ y+\frac { { y }^{ 2 } }{ 2 } \right] }_{ -1 }^{ 3 }=\pi \left[ \left( 3+\frac { 9 }{ 2 } \right) -\left( -1+\frac { 1 }{ 2 } \right) \right] \)
\(=\pi \left[ \left( \frac { 6+9 }{ 2 } \right) -\left( -\frac { 1 }{ 2 } \right) =\pi \left( \frac { 15 }{ 2 } +\frac { 1 }{ 2 } \right) =\left( \frac { 16 }{ 2 } \right) \right] \)
V = 8\(\pi\)