The evaluation of an indefinite integral of the form
∫ u(x) v(x)dx
becomes very simple, when u is a polynomial function of x and v(x) can be easily integrated successively.
It is accomplished by a formula called Bernoulli’s formula. This formula is actually an extension of the formula of integration by parts. To derive the formula, we use the following notation:
∫ udv = uv-u'v_{1}+u''v_{2 - }+u'''v_{3 +...........}
The above result is called the Bernoulli’s formula for integration of product of two functions.
Note :
Since u is a polynomial function of x , the successive derivative u(m)will be zero for some positive integer m and so all further derivatives will be zero only.
Hence the right-hand-side of the above formula contains a finite number of terms only.
Evaluate the following.
Problem 1 :
Solution :
= e^{-2} (1-3/2 - 6/4 - 6/16) + 3/8
= e^{-2} (1-3/2 - 3/2 - 3/8) + 3/8
= e^{-2} (1-6/2 - 3/8) + 3/8
= e^{-2} (8-24-3)/8 + 3/8
= -19e^{-2}/8 + 3/8
Problem 2 :
Solution :
= π/4 [(-cos 3π/4)/3] + sin(3π/4)/9 - 0 - 0
= π/4 [(1/3√2)] + 1/9√2
= π/12√2 + 1/9√2
= (1/√2)[π/12 + 1/9]
Problem 3 :
Solution :
Using Bernoulli's theorem,
u = t, dv = e^{t} dt
u' = 1, v = e^{t} dt and v_{1} = e^{t}
= [te^{t }- 1e^{t}] limits 0 to π/4
= [π/4e^{π/4 }^{ }- 1e^{π/4 }+ 1]
= e^{π/4}(π/4^{ }^{ }- 1)^{ }+ 1
Problem 4 :
Solution :
Let u = x^{2 }and dv = cos 2x
u' = 2x, u'' = 2
v = sin 2x/2, v_{1} = -cos2x/4, v_{2} = -sin2x/8
= [x^{2}(sin 2x/2) - 2x (-cos2x/4) + 2(-sin2x/8)] 0 to π/2
By applying the limit, we get
= -π/4
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