Integral by Parts: Definite Integral
How to solve the definite integral of a product by using the integral by parts: formula, 2 examples, and their solutions.
Formula
The integral of a product, uv', can be solved
by using this formula.
∫ab uv' dx = [uv]ab - ∫ab u'v dx
How to Choose u and v'
The priority of choosing u and v'
is the same as
the case of solving an indefinite integral.
For a product of two functions in an integral,
one is u, and the other is v'.
To choose u and v' properly,
remember this order of uv':
[u]
logarithmic
polynomial
trigonometric,
exponential
[v'].
Set u = (upper function) and v' = (lower function).
u is the function
whose integral is complex.
v' is the function
whose integral is simple.
Example∫0π/3 (x sin x) dx
x sin x is the product of x and sin x.
So solve this
by using integral by parts.
The order of uv' is
[u]
logarithmic
polynomial (x)
trigonometric (sin x)
exponential
[v'].
So set
u = x and v' = sin x.
Write u = x.
Differentiate both sides.
Then u' = 1.
Derivative of a Polynomial
Write v' = sin x
next to u' = 1.
Integrate both sides.
Then v = -cos x.
Integral of sin x
Write this above v' = sin x.
u = x, v = -cos x
u' = 1
Then the given integral is equal to,
uv, x⋅(-cos x)
from 0 to π/3
minus
integral from 0 to π/3,
u'v, 1⋅(-cos x) dx.
x⋅(-cos x) = -x cos x
-∫0π/3 1⋅(-cos x) dx = +∫0π/3 cos x dx
Put 0 and π/3
into -x cos x.
Then -(π/3)⋅(cos π/3) - (0⋅cos 0).
Solve the integral.
Definite Integral: How to Solve
The integral of cos x is
sin x.
So -(π/3)⋅(cos π/3) - (0⋅cos 0) + [sin x]0π/3.
cos π/3 = 1/2
So -(π/3)⋅(cos π/3) = -(π/3)⋅(1/2).
Cosine Values of Commonly Used Angles
Put 0 and π/3
into sin x.
Then sin π/3 - sin 0.
-(π/3)⋅(1/2) = -π/6
sin π/3 = √3/2
sin 0 = 0
Sine Values of Commonly Used Angles
-π/6 + [√3/2 - 0] = √3/2 - π/6
So
√3/2 - π/6
is the answer.
Example∫1e (ln x)2 dx
(ln x)2 is the product of (ln x)2 and 1.
So solve this
by using integral by parts.
The order of uv' is
[u]
logarithmic [(ln x)2]
polynomial (1)
trigonometric
exponential
[v'].
So set
u = (ln x)2 and v' = 1.
Write u = (ln x)2.
Differentiate both sides.
Then u' = 2(ln x)⋅(1/x).
Derivative of a Composite Function
Derivative of ln x
Write v' = 1
next to u' = 2(ln x)1⋅(1/x).
Integrate both sides.
Then v = x.
Integral of a Polynomial
Write this above v' = 1.
u = (ln x)2, v = 2(ln x)⋅(1/x)
u' = x
Then the given integral is equal to,
uv, (ln x)2⋅x
from 1 to e
minus
integral from 1 to e,
u'v, 2(ln x)⋅(1/x)⋅x dx.
Put 1 and e
into (ln x)2⋅x.
Then (ln e)2⋅e - (ln 1)2⋅1.
Take 2 out from the integral.
And cancel (1/x) and x.
Then -∫1e 2(ln x)⋅(1/x)⋅x dx = -2⋅∫1e ln x dx
So (ln e)2⋅e - (ln 1)2⋅1 - 2⋅∫1e ln x dx.
12⋅e = e
Put 1 and e
into x ln x - x.
Then -2[e ln e - e - (1 ln 1 - 1)].
e ln e = e⋅1
1 ln 1 = 1⋅0
e⋅1 - e = 0
(1⋅0 - 1) = (-1)
0 - (-1) = 0 + 1 = +1
-2[+1] = -2
So
e - 2
is the answer.