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In what follows, c is a constant of integration and can take any contant value.
1 - Integral of k f(x).
ò k f(x) dx = k ò f(x) dx
Example: Evaluate the integral
ò 5 sinx dx
Solution:
According to the above property
ò 5 sinx dx = 5ò sinx dx
ò sinx dx is given by 2.1 in table of integrals, hence
ò 5 sinx dx = -5cos x + c
2 - Integral of Sum of Functions.
ò [f(x) + g(x)] dx = òf(x) dx + òg(x) dx
Example: Evaluate the integral
ò [x + e x] dx
Solution:
According to the above property
ò [x + e x] dx = òx dx + òe x dx
ò x dx is given by 1.3 and òe x dx
by 4.1 in table of integrals, hence
ò [x + e x] dx = x 2 / 2 + e x + c
3 - Integral of Difference of Functions.
ò [f(x) - g(x)] dx = òf(x) dx - òg(x) dx
Example: Evaluate the integral
ò [2 - 1/x] dx
Solution:
According to the above property
ò [2 - 1/x] dx = ò2 dx - ò (1/x) dx
ò 2 dx is given by 1.2 and ò(1/x) dx
by 1.4 in table of integrals, hence
ò [2 - 1/x] dx = 2x - ln |x| + c
3 - Integration by Substitution.
ò [f(u) du/dx] dx = òf(u) du
Example: Evaluate the integral
ò (x 2 - 1) 20 2x dx
Solution:
Let u = x 2 - 1, du/dx = 2x and the given integral can be written as
ò (x 2 - 1) 20 2x dx
= ò u 20 (du/dx) dx
= ò u 20 du according to above property
= u 21 / 21 + c = (x 2 - 1) 21 / 21 + c
3 - Integration by Parts.
ò f(x) g '(x) dx = f(x) g(x) - òf '(x) g(x) dx
Example: Evaluate the integral
ò x cos x dx
Solution:
Let f(x) = x and g ' (x) = cos x which gives
f ' (x) = 1 and g(x) = sin x
From integration by parts formula above,
ò x cos x dx = x sin x - ò1 sin x dx
= x sin x + cos x + c
Exercises: Use the table of integrals and the properties above to evaluate the following integrals. [Note that you may need to use more than one of the above properties for one integral].
1. ò (1/2) ln x dx
2. ò [sin x + x 5] dx
3. ò [sinh x - 3] dx
4. ò -x sin x dx
Answers to Above Exercises
1. [x ln x] / 2 - x / 2
2. x 6 / 6 - cos x
3. cosh x - 3x
4. x cos x - sin x
More references on
integrals and their applications in calculus.
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