Applying integration by parts, we get
\(\int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ n } } dx={ \left[ { x }^{ n }(-{ e }^{ -x }) \right] }_{ 0 }^{ \infty }-\int _{ 0 }^{ \infty }{ (-{ e }^{ -x }) } ({ nx }^{ n-1 })dx=n\int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ n-1 }dx } \)
\(Let\quad { I }_{ n }=\int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ n }dx.\ Then,\ { I }_{ n }={ nI }_{ n-1 } } \)
So, we get I_n= n(n−1) I_n−2.
Proceeding in this way, we get ultimately,
I_n = n(n-1)(n-2)...(2)(1)I₀.
But, \({ I }_{ 0 }=\int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ 0 }dx } ={ \left( -{ e }^{ -x } \right) }_{ 0 }^{ \infty }=0+1=1.\) So, we get n = n(n-1)(n-2)(2)(1) = n!
Hence, we get
\(\int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ n }dx=n! } \), where n is a nonnegative integer.

The equation of the latus-rectum is x = a. It intersects the parabola at the points L(a, 2a) and L₁ (a, −2). The required area is sketched. By symmetry, the required area A is twice the area bounded by the portion of the parabola

y = 2\(\sqrt a \sqrt x\), x -axis, x = 0 and x = a.
Hence, by taking vertical strips, we get
\(A=2\int _{ 0 }^{ a }{ ydx=2\int _{ 0 }^{ a }{ 2\sqrt { a } \sqrt { x } dx=4 } \sqrt { a } } { \left[ \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ a }\)
\(=4\sqrt { a } \times \frac { 2 }{ 3 } { a }^{ \frac { 3 }{ 2 } }=\frac { 8{ a }^{ 2 } }{ 3 } \)

The equation of the parabola is ( y + 2)² = −(x−9). The parabola crosses the y- axis at (0, −5) and (0, 1). The vertex is at (9, −2) and the axis of the parabola is y = −2. The required area is sketched
Viewing in the positive direction of x-axis, and making horizontal strips, the required area A is given by

\(A=\int _{ -5 }^{ 1 }{ xdy } =\int _{ -5 }^{ 1 }{ (5-4y-{ y }^{ 2 })dy={ \left[ 5y-2{ y }^{ 2 }-\frac { { y }^{ 3 } }{ 3 } \right] }_{ -5 }^{ 1 }=\frac { 8 }{ 3 } -\left( -\frac { 100 }{ 3 } \right) =36 } \)

The required area is sketched. One portion of the region lies above the x−axis between x = 0 and x = \(\pi\), and the other portion lies below x−axis between x = \(\pi\) and x = 2\(\pi\). So, the required area is given by 

If the region in the first quadrant bounded by the circle x² + y² = r², the x-axis, the lines x = r − h and x = r is revolved about the x-axis, then the solid generated is a spherical cap of height h cut of from a sphere of radius r. Hence, the required volume is given by

\(V=\pi \int _{ r-h }^{ r }{ { y }^{ 2 }dx=\pi \int _{ r-h }^{ r }{ \left( { r }^{ 2 }-{ x }^{ 2 } \right) dx } =\pi } { \left( { r }^{ 2 }x-\frac { { x }^{ 3 } }{ 3 } \right) }_{ r-h }^{ r }\)
\(=\pi \left( { r }^{ 2 }(r-(r-h)-\frac { \left( { r }^{ 3 }-{ (r-h) }^{ 3 } \right) }{ 3 } \right) =\pi \left( { r }^{ 2 }h-\frac { \left( { r }^{ 3 }-\left( { r }^{ 3 }-3{ r }^{ 2 }h+3{ rh }^{ 2 }-{ h }^{ 3 } \right) \right) }{ 3 } \right) \)
\(\\ =\pi \left( \frac { { 3rh }^{ 2 }-{ h }^{ 3 } }{ 3 } \right) =\frac { 1 }{ 3 } \pi { h }^{ 2 }(3r-h)\)

The ellipse is symmetric about both the axes. The major axis lies along x-axis. The region to be revolved is sketched
Hence, the required volume is given by
\(V=\pi \int _{ -a }^{ a }{ { y }^{ 2 }dx } =\pi \int _{ -a }^{ a }{ \frac { { b }^{ 2 } }{ { a }^{ 2 } } \left( { a }^{ 2 }-{ x }^{ 2 } \right) dx } \)
\(=\frac { 2\pi { b }^{ 2 } }{ { a }^{ 2 } } \int _{ 0 }^{ a }{ ({ a }^{ 2 }-{ x }^{ 2 })dx } \) since the integrand is an even function
\(=\frac { { 2\pi b }^{ 2 } }{ { a }^{ 2 } } { \left( { a }^{ 2 }x-\frac { { x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ a }=\frac { 2\pi { b }^{ 2 } }{ 3 } ={ \left( { a }^{ 2 }-\frac { { a }^{ 3 } }{ 3 } \right) }_{ 0 }^{ a }=\frac { 2\pi { b }^{ 2 } }{ 3 } \left( \frac { { 2a }^{ 3 } }{ 3 } \right) =\frac { 4\pi { ab }^{ 2 } }{ 3 } \)

The region is sketched. It lies above the x − axis. Hence, the required area is given by 

\(A=\int _{ -1 }^{ 3 }{ ydx } =\int _{ -1 }^{ 3 }{ \left( \frac { 30-6x }{ 5 } \right) dx={ \left( \frac { 30x-3{ x }^{ 2 } }{ 5 } \right) }_{ -1 }^{ 3 } } \)
\(=\left( \frac { 90-27 }{ 5 } \right) -\left( \frac { -30-3 }{ 5 } \right) =\frac { 96 }{ 5 } \)

Given equation of line is 3x - 2y + 6 = 0
2y = 3x + 6 \(\Rightarrow\) y = \(\frac{3x+6}{2}\)

\(\therefore Area=\int _{ -3 }^{ -2 }{ -ydx+ydx } +\int _{ -2 }^{ 1 }{ ydx } \)
[\(\because\) the Area is below the x - axis]

\(=\frac { -1 }{ 2 } \int _{ -3 }^{ -2 }{ (3x+6)dx+\frac { 1 }{ 2 } \int _{ -2 }^{ 1 }{ (3x+6)dx } } \)
\(={ \left[ \frac { { 3x }^{ 2 } }{ 2 } +6x \right] }_{ -3 }^{ -2 }+\frac { 1 }{ 2 } { \left[ \frac { { 3x }^{ 2 } }{ 2 } +6x \right] }_{ -2 }^{ 1 }\)
\(=-\frac { 1 }{ 2 } \left[ \left( \frac { 12 }{ 2 } -12 \right) -\left( \frac { 27 }{ 2 } -18 \right) \right] +\frac { 1 }{ 2 } \left[ \left( \frac { 3 }{ 2 } +6 \right) -\left( \frac { 12 }{ 2 } -12 \right) \right] \)
\(=-\frac { 1 }{ 2 } \left[ (-6)-\left( \frac { 27-36 }{ 2 } \right) \right] +\frac { 1 }{ 2 } \left[ \left( \frac { 3+12 }{ 2 } \right) -(-6) \right] \)
\(=-\frac { 1 }{ 2 } \left[ -6+\frac { 9 }{ 2 } \right] +\frac { 1 }{ 2 } \left[ \frac { 15 }{ 2 } +6 \right] \)
\(\\ =-\frac { 1 }{ 2 } \left[ \frac { -3 }{ 2 } \right] +\frac { 1 }{ 2 } \left[ \frac { 27 }{ 2 } \right] =\frac { 3 }{ 4 } +\frac { 27 }{ 4 } =\frac { 30 }{ 4 } =\frac { 15 }{ 2 } \)
\(\therefore\)  A = 7.5 sq.units

Solution the region is sketched. It lies below the x − axis. Hence, the required area is given by

\(A=\left| \int _{ -2 }^{ 3 }{ ydx } \right| =\left| \int _{ -2 }^{ 3 }{ \left( \frac { 7x-35 }{ 5 } \right) dx } \right| \)
\(=\frac { 1 }{ 5 } \left| { \left( 7\left( \frac { { x }^{ 2 } }{ 2 } \right) -35x \right) }_{ -2 }^{ 3 } \right| \)
\(=\frac { 1 }{ 5 } \left| \left( \left( \frac { 63 }{ 2 } \right) -105 \right) -(84) \right| =\frac { 63 }{ 2 } \)

By revolving the upper semicircular region enclosed between the circle x² + y² = a² and the x-axis, we get a sphere of radius a.

The boundaries of the region are y = \(\\ \\ \\ \\ \\ \\ \\ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } \) x-axis, the lines x = −a and x = a. Hence, the volume of the sphere is given by
\(v=\pi \int _{ -a }^{ a }{ { y }^{ 2 }dx=\pi } \int _{ -a }^{ a }{ \left( { a }^{ 2 }-{ x }^{ 2 } \right) } dx\) 
\(=2\pi \int _{ 0 }^{ a }{ ({ a }^{ 2 }-{ x }^{ 2 })dx } \) since the integrand (a²-x²) is an even function
\(=2\pi { \left( { a }^{ 2 }x-\frac { { x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ a }=2\pi \left( { a }^{ 3 }-\frac { { a }^{ 3 } }{ 3 } \right) =\frac { 4 }{ 3 } { \pi a }^{ 3 }\)