Question

Suppose f is such that \[f\left( { - x} \right) = - f\left( x \right)\;\]for every real x and \[\int_0^1 {f\left( x \right)dx} = 5\], then find the value of \[\int_{ - 1}^0 {f\left( t \right)dt} \].
A. 10
B. 0
C. 5
D. -5

Answer 1

Hint: We will determine the nature of f. It means we will find whether f is an odd function or an even function by using given relation. Then we apply the property of definite integral to calculate the value of \[\int_{ - 1}^1 {f\left( x \right)dx} \]. Then break the interval of the integration and substitute the value of \[\int_0^1 {f\left( x \right)dx} \] to the value of \[\int_{ - 1}^0 {f\left( t \right)dt} \].



Formula Used:Definite integral property:
\[\int_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}{0\,{\rm{if}}\,f\left( { - x} \right) = - f\left( x \right)}\\{2\int\limits_0^a {f\left( x \right)dx\,} {\rm{if}}\,f\left( { - x} \right) = f\left( x \right)}\end{array}} \right.\]
\[\int_b^a {f\left( x \right)dx} = \int_b^c {f\left( x \right)dx} + \int_c^a {f\left( x \right)dx} \] where \[b < c < a\].



Complete step by step solution:Given that \[f\left( { - x} \right) = - f\left( x \right)\;\].
Thus \[f\left( x \right)\;\] is an odd function.
According the property of definite integral \[\int_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}{0\,{\rm{if}}\,f\left( { - x} \right) = - f\left( x \right)}\\{2\int\limits_0^a {f\left( x \right)dx\,} {\rm{if}}\,f\left( { - x} \right) = f\left( x \right)}\end{array}} \right.\]
That is\[\int_{ - a}^a {f\left( x \right)dx} = 0\]
Now putting a = 1
\[\int_{ - 1}^1 {f\left( x \right)dx} = 0\]
Apply the property \[\int_b^a {f\left( x \right)dx} = \int_b^c {f\left( x \right)dx} + \int_c^a {f\left( x \right)dx} \] where a = 1, b = -1 and c = 0
\[\int_{ - 1}^1 {f\left( x \right)dx} = \int_{ - 1}^0 {f\left( x \right)dx} + \int_0^1 {f\left( x \right)dx} = 0\]
Now substitute \[\int_0^1 {f\left( x \right)dx} = 5\]
\[ \Rightarrow \int_{ - 1}^0 {f\left( x \right)dx} + 5 = 0\]
Subtract 5 from both sides:
\[ \Rightarrow \int_{ - 1}^0 {f\left( x \right)dx} = - 5\]
Now applying the property \[\int {f\left( x \right)dx} = \int {f\left( t \right)dt} \]:
\[ \Rightarrow \int_{ - 1}^0 {f\left( t \right)dt} = - 5\]



Option ‘D’ is correct



Note: Students often make mistake when they solve this type question. They evaluate the value of \[f\left( x \right)\] from relation \[f\left( { - x} \right) = - f\left( x \right)\;\] and substitution in \[\int_0^1 {f\left( x \right)dx} = 5\]. Then they try to solve it. But it is incorrect way. To solve this type of definite integral, we have to use definite integral property.