FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 3.75INTEGRATION BY PARTS
Integration of a Product:
Let u and v 1 be differential functions of x.
Then dxd(uv ) v u 1 =dudx 1 + dvdx^1 (from diff. calculus)
Now integrating both sides w.r.t. x
We get 1 1 1
uv duv dx u dxdv
dx du
=^ +^
(^)
or,^111
u dx uvdv duv dx
dx dx
(^) = −
(transposing)
Taking dvdx^1 =v then v 1 = vdx
The above result may be written as
(uv)dx u vdx du vdx dx
dx
= −^
(^)
It states integral of product of two functions = 1st function (unchanged) × int. of 2nd – integral of
(diff. 1st × int. of 2nd.).
Note : Care should be taken to choose properly the first function, i.e., the function not to be integrated.
Example 157:Evaluate xe dxx
xe dxx (here ex is taken as second function)
x e dxx dx e dx. dx x.e 1e dx xe e cx x x x x
dx
= −^ = − = − +
Note : If ex be taken as first function, integral becomes e xdxx
x d x x x^2 x x^2=e xdx− (^) dxe. xdx dx e. = 2 − e. dx 2
Now to find the value of
x x^2
e. 2 dx becomes complicated. So x is taken as first function for easy solution.