Question

If $y = f\left( x \right)$ makes positive intercept of $2$ and $0$ unit on x and y axes and encloses an area of $\dfrac{3}{4}$ square unit within the axes, then what is the value of $\int\limits_0^2 {xf'\left( x \right)dx} $ ?
A. $\dfrac{3}{4}$
B. $1$
C. $\dfrac{5}{4}$
D. $\dfrac{7}{4}$

Answer 1

Hint:In the given question we are asked to evaluate the value of $\int\limits_0^2 {xf'\left( x \right)dx} $ , when certain other values are also given. Solve this integral using integration by parts and make use of the other values available to calculate its value.

Formula used:The integral \[\int {u.vdx} \] , where $u$ and $v$ are both functions of $x$ , is calculated using integration by parts as:
$\int {I.IIdx} = I\int {IIdx - \int {\left[ {\left( {\dfrac{{dI}}{{dx}}} \right)\int {IIdx} } \right]dx} } $
Here, $u$ and $v$ are considered as $I$ and $II$ functions as per our convenience.

Complete step by step Solution:
It is given that $y = f\left( x \right)$ encloses an area of $\dfrac{3}{4}$ square unit within the axes.
Thus, $\int_0^2 {f\left( x \right)dx = \dfrac{3}{4}} $ … (1)
Now, we are asked to evaluate the value of $\int\limits_0^2 {xf'\left( x \right)dx} $ .
Applying integration by parts and consider $x$ as the first function and $f'\left( x \right)$ as the second function,
$\int\limits_0^2 {xf'\left( x \right)dx} = x\int\limits_0^2 {f'\left( x \right)dx - \int\limits_0^2 {f\left( x \right)dx} } $
We know that $\int {f'\left( x \right)dx = f\left( x \right)} $ . Therefore, substituting that in the above equation,
\[\int\limits_0^2 {xf'\left( x \right)dx} = \left[ {xf\left( x \right)} \right]_0^2 - \int\limits_0^2 {f\left( x \right)dx} \]
Using (1),
\[\int\limits_0^2 {xf'\left( x \right)dx} = \left( {2f\left( 2 \right) - 0} \right) - \dfrac{3}{4}\]
Also, we are provided that $y = f\left( x \right)$ makes a positive intercept of $2$ units on x-axis, this means, that $f(2) = 0$ , hence, substituting that, we get:
\[\int\limits_0^2 {xf'\left( x \right)dx} = - \dfrac{3}{4}\]

Therefore, the correct option is (D).

Note: While using integration by parts to calculate the value of an integral, we are given a product of two functions of $x$ in the integral. The selection of first and second functions to be considered while using integration by parts can be made easier using ILATE rule.