Question
Answers

If function f(x) is an odd function, then find the value of following integral:
\[\int_{-a}^{a}{f(x)dx}\]
A) \[2\int_{0}^{a}{f(x)dx}\]
B) \[\int_{0}^{a}{f(x)dx}\]
C) 0
D) \[\int_{0}^{a}{f(a-x)dx}\]

Answer Verified Verified
Hint: Integrate the given expression directly and use the concept given by ‘the integration of an Odd function is always even.’

We will write the expression, of which we have to find the value and assume it as L,
\[L=\int_{-a}^{a}{f(x)dx}\]……………………………… (1)
Suppose,
\[\int{f(x)}dx=F(x)\]…………………………………. (2)
To proceed further we should know the formula of definite integration given below,
Formula:
\[\int\limits_{a}^{b}{f(x)dx}=\left[ F(x) \right]_{a}^{b}\]
By using the formula and equation (2) we can write equation (1) as,
\[\therefore L=\left[ F(x) \right]_{-a}^{a}\]
To proceed further we should know how to substitute the limits in f(x) which is shown below,
\[\left[ F(x) \right]_{a}^{b}=F(b)-F(a)\]
By using above formula we can write ‘L’ as shown below,
\[\therefore L=\left[ F(x) \right]_{-a}^{a}=F(a)-F(-a)\]
\[\therefore L=F(a)-F(-a)\]………………………………………… (3)
To solve the above equation further we should know the concept of integration of odd functions given below,
Concept:
Integration of an Odd function is always even.
As F(x) is the integration of f(x) therefore it is an even function and therefore we should know how the even functions are expressed.
As F(x) is an even function therefore we should know the concept given below to proceed further in the solution,
Concept:
If F(x) is an even function then,
F(-x) = F(x), For all values of x.
By using the above concept we can write the equation (3) as,
\[\therefore L=F(a)-F(a)\]
\[\therefore L=0\]………………………………………. (4)
Therefore from equation (4) and equation (1) we can write,
\[\therefore L=\int_{-a}^{a}{f(x)dx}=0\]
Therefore if f(x) is an odd function then \[\int_{-a}^{a}{f(x)dx}\] is equal to ‘0’
Therefore option (c) is the correct answer.

Note: You should know that, if F(x) is an even function then, F(-x) = F(x) to get the right answer.
Bookmark added to your notes.
View Notes
×