Question

What is the value of \[\int\limits_0^{2a} {f\left( x \right)dx} \]?
A. \[2\int_0^a {f\left( x \right)dx} \]
B. 0
C. \[\int\limits_0^a {f\left( x \right)dx} + \int\limits_0^a {f\left( {2a - x} \right)dx} \]
D. \[\int\limits_0^a {f\left( x \right)dx} + \int\limits_0^{2a} {f\left( {2a - x} \right)dx} \]

Answer 1

Hint: First we will apply the definite integral property that is \[\int_b^a {f\left( x \right)dx} = \int_b^c {f\left( x \right)dx} + \int_c^a {f\left( x \right)dx} \]. They we apply substitution method in the second integration and replace the variable with x to get required result.



Formula Used:The property of definite integration:
\[\int_b^a {f\left( x \right)dx} = \int_b^c {f\left( x \right)dx} + \int_c^a {f\left( x \right)dx} \], where \[b < c < a\].
\[\int_b^a {f\left( u \right)du} = \int_b^a {f\left( x \right)dx} \]



Complete step by step solution:The given definite integration is \[\int\limits_0^{2a} {f\left( x \right)dx} \].
We know that \[0 < a < 2a\]
Now applying the property of the definite integration \[\int_b^a {f\left( x \right)dx} = \int_b^c {f\left( x \right)dx} + \int_c^a {f\left( x \right)dx} \]:
Assume that \[b = 0\], \[c = a\], \[a = 2a\]:
\[\int\limits_0^{2a} {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} + \int\limits_a^{2a} {f\left( x \right)dx} \] …..(i)
Assume that, \[x = 2a - u\]
Differentiate both sides with respect to x:
\[dx = - du\]
When \[x = a\], then \[a = 2a - u \Rightarrow u = a\]
When \[x = 2a\], then \[2a = 2a - u \Rightarrow u = 0\]
Now applying substitution method in the second integration of (i)
\[\int\limits_0^{2a} {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} - \int\limits_a^0 {f\left( {2a - u} \right)du} \]
Now applying the property \[\int_b^a {f\left( x \right)dx} = - \int_a^b {f\left( x \right)dx} \]:
\[\int\limits_0^{2a} {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} + \int\limits_0^a {f\left( {2a - u} \right)du} \]
Now replacing u by x:
\[\int\limits_0^{2a} {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} + \int\limits_0^a {f\left( {2a - x} \right)dx} \]



Option ‘C’ is correct



Note: Students often stuck in the step \[\int\limits_0^{2a} {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} + \int\limits_0^a {f\left( {2a - u} \right)du} \]. They forgot about the property \[\int_b^a {f\left( u \right)du} = \int_b^a {f\left( x \right)dx} \]. This means we can replace a variable. We replace the variable u with x of the definite integration \[\int\limits_0^{2a} {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} + \int\limits_0^a {f\left( {2a - u} \right)du} \].