How to use integration by parts in definite integral?
Answers
Answer:
Integration by parts is a technique for performing indefinite integration intudv or definite integration int_a^budv by expanding the differential of a product of functions d(uv) and expressing the original integral in terms of a known integral intvdu. A single integration by parts starts with
d(uv)=udv+vdu,
(1)
and integrates both sides,
intd(uv)=uv=intudv+intvdu.
(2)
Rearranging gives
intudv=uv-intvdu.
(3)
For example, consider the integral intxcosxdx and let
u=x dv=cosxdx
(4)
du=dx v=sinx,
(5)
so integration by parts gives
intxcosxdx = xsinx-intsinxdx
(6)
= xsinx+cosx+C,
(7)
where C is a constant of integration.
The procedure does not always succeed, since some choices of u may lead to more complicated integrals than the original. For example, consider again the integral intxcosxdx and let
u=cosx dv=xdx
du=-sinxdx v=1/2x^2,
(8)
giving
intxcosxdx = 1/2x^2cosx-1/2intx^2(-sinx)dx
(9)
= 1/2x^2cosx+1/2intx^2sinxdx,
(10)
which is more difficult than the original (Apostol 1967, pp. 218-219).
Integration by parts may also fail because it leads back to the original integral. For example, consider intx^(-1)dx and let
u=x dv=x^(-2)dx
du=dx v=-x^(-1),
(11)
then
intx^(-1)dx=-1-int(-x^(-1))dx+C=intx^(-1)dx+C-1,
(12)
which is same integral as the original (Apostol 1967, p. 219).
The analogous procedure works for definite integration by parts, so
int_a^budv=[uv]_a^b-int_a^bvdu,
(13)
where [f]_a^b=f(b)-f(a).
Integration by parts can also be applied n times to intf^((n))(x)g(x)dx:
u=g(x) dv=f^((n))(x)dx
du=g^'(x)dx v=f^((n-1))(x).
(14)
Therefore,
intf^((n))(x)g(x)dx=g(x)f^((n-1))(x)-intf^((n-1))(x)g^'(x)dx.
(15)
But
intf^((n-1))(x)g^'(x)dx=g^'(x)f^((n-2))(x)-intf^((n-2))(x)g^('')(x)dx
(16)
intf^((n-2))(x)g^('')(x)dx=g^('')(x)f^((n-3))(x)-intf^((n-3))(x)g^((3))(x)dx,
(17)
so
intf^((n))(x)g(x)dx=g(x)f^((n-1))(x)-g^'(x)f^((n-2))(x)
+g^('')(x)f^((n-3))(x)-...+(-1)^nintf(x)g^((n))(x)dx.
(18)
Now consider this in the slightly different form intf(x)g(x)dx. Integrate by parts a first time
u=f(x) dv=g(x)dx
du=f^'(x)dx v=intg(x)dx,
(19)
so
intf(x)g(x)dx=f(x)intg(x)dx-int[intg(x)dx]f^'(x)dx.
(20)
Now integrate by parts a second time,
u=f^'(x) dv=intg(x)dx
du=f^('')(x)dx v=intintg(x)(dx)^2,
(21)
so
intf(x)g(x)dx=f(x)intg(x)dx-f^'(x)intintg(x)(dx)^2+int[intintg(x)(dx)^2]f^('')(x)dx.
(22)
Repeating a third time,
intf(x)g(x)dx=f(x)intg(x)dx-f^'(x)intintg(x)(dx)^2+f^('')(x)intintintg(x)(dx)^3-int[intintintg(x)(dx)^3]f^(''')(x)dx.
(23)
Therefore, after n applications,
intf(x)g(x)dx=f(x)intg(x)dx-f^'(x)intintg(x)(dx)^2+f^('')(x)intintintg(x)(dx)^3-...+(-1)^(n+1)f^((n))(x)int...int_()_(n+1)g(x)(dx)^(n+1)+(-1)^nint[int...int_()_(n+1)g(x)(dx)^(n+1)]f^((n+1))(x)dx.
(24)
If f^((n+1))(x)=0 (e.g., for an nth degree polynomial), the last term is 0, so the sum terminates after n terms and
intf(x)g(x)dx=f(x)intg(x)dx-f^'(x)intintg(x)(dx)^2+f^('')(x)intintintg(x)(dx)^3-...+(-1)^(n+1)f^((n))(x)int...int_()_(n+1)g(x)(dx)