Question

Let $F:R\to R$ be a thrice differential function. Suppose that F(1) = 0, F(3) = -4 and F’(x) < 0 for all $x\in \left( 1,2,3 \right)$. Let f(x) =x F(x) for all $x\in R$ . If $\int\limits_{1}^{3}{{{x}^{2}}F'(x)dx=-12}$ and \[\int\limits_{1}^{3}{{{x}^{3}}F(x)dx=40}\], then correct expression(s) is (are)
(a) 9f’(3) + f’(1) - 32 = 0
(b) $\int\limits_{1}^{3}{xF(x)dx=12}$
(c) 9f’(3) - f’(1) + 32 = 0
(d) $\int\limits_{1}^{3}{xF(x)dx=-12}$

Answer 1

Hint: To solve this question, what we will do first is, we will first use ILATE to solve the integral. And then after some simplification, we will substitute values of F for x = 1 and x = 3 and after that using condition f(x) =x F(x) and some calculation, we will end up with answer.

Complete step by step answer:
We know that, if we have two functions of x in multiplication for integration, then we use concept of ILATE.
ILATE stands for
I - Inverse function,
L - Logarithmic function,
A - Algebraic function,
T - Trigonometric function,
E - Exponential function.
Now, if we have $\int{u(x)v(x)dx}$ , then priority of selection of function u ( x ) decreases from Inverse function to Exponential and value of integration is $\int{u(x)v(x)dx}=u(x)\int{v(x)dx-\int{\left( \dfrac{d}{dx}u(x)\int{v(x)dx} \right)}}dx$
Also, To do such question, we must know some properties of definite integration and indefinite integration..
One of the most important property of definite integration is \[\int\limits_{a}^{b}{f(x)dx=F(a)-F(b)}\] ,where F is the integration of f ( x ) and a is lower limit and b is upper limit. That is for example we have $f(x)=x$ and we have to find integration of f ( x ) = x from lower limit 2 to upper limit 3, then
 \[{{\int\limits_{2}^{3}{xdx=\left[ \dfrac{{{x}^{2}}}{2} \right]}}_{2}}^{3}\]
\[\int\limits_{2}^{3}{xdx=}\dfrac{{{(3)}^{2}}}{2}-\dfrac{{{(2)}^{2}}}{2}=2.5\], as integration of x is $\dfrac{{{x}^{2}}}{2}$ because $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n}+C}$
So, we are provided in question that, $\int\limits_{1}^{3}{{{x}^{2}}F'(x)dx=-12}$
Taking ${{x}^{2}}$ as first function and F’(x) as second function, then by using ILATE, we get
\[\int\limits_{1}^{3}{{{x}^{2}}F'(x)dx}=\left[ {{x}^{2}}\int{F'(x)dx} \right]_{1}^{3}-\int\limits_{1}^{3}{\left( \dfrac{d}{dx}{{x}^{2}}\int\limits_{1}^{3}{F'(x)} \right)dx}\]
Now, on solving we get
\[\int\limits_{1}^{3}{{{x}^{2}}F'(x)dx}=\left[ {{x}^{2}}(F(x))_{1}^{3}-\int\limits_{1}^{3}{\left( 2x(F(x))_{1}^{3} \right)dx} \right]\]
\[\int\limits_{1}^{3}{{{x}^{2}}F'(x)dx}=\left[ {{x}^{2}}(F(3)-F(1))-\int\limits_{1}^{3}{\left( 2x(F(3)-F(1)) \right)dx} \right]\]
We are given that F(1) = 0, F(3) = -4, so
\[\int\limits_{1}^{3}{{{x}^{2}}F'(x)dx}=\left[ {{3}^{2}}F(3)-{{1}^{2}}F(1)-2\int\limits_{1}^{3}{\left( xF(x) \right)dx} \right]\]
As, in question it is given that f(x) =x F(x), we can write above expression as
\[=-9.4-1.0-2\int\limits_{1}^{3}{f(x)dx}=-12\]
\[=-36-2\int\limits_{1}^{3}{f(x)dx}=-12\]
\[=-2\int\limits_{1}^{3}{f(x)dx}=-12+36\]
\[=-2\int\limits_{1}^{3}{f(x)dx}=24\]
On solving, we get
\[=\int\limits_{1}^{3}{f(x)dx}=-12\]
As, f(x) =x F(x)
So, we can say that
\[=\int\limits_{1}^{3}{xF(x)dx}=-12\]
Hence, option ( d ) is correct.

Note:
While integrating the definite integral, always use the correct formula to evaluate the integration and always try to skip calculation error as it may change the answer of the solution or may make the solution more complex. One must know all the properties and formula of indefinite and definite integrals.