Question

The value of the integral $\int_{4}^{10}{\dfrac{\left[ {{x}^{2}} \right]}{\left[ {{x}^{2}}-28x+196 \right]+\left[ {{x}^{2}} \right]}dx}$, where $\left[ x \right]$ denotes the greater integer less than or equal to x, is: -
(a) 7
(b) 6
(c) 3
(d) $\dfrac{1}{3}$

Answer 1

Assume the value of the integral as ‘I’. Write ${{x}^{2}}-28x+196$ in the form of ${{\left( x-a \right)}^{2}}$. Assume this expression of I as equation (i). Now, apply the formula given as: - $I=\int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left( a+b-x \right)dx}$, where f (x) is the function inside the integral sign and ‘a’ and ‘b’ are the given lower and upper limits respectively. Assume this converted form as equation(ii). Add both the equations and cancel the like terms to finally evaluate the limit of integration.

Complete step-by-step answer:
We have to find the value of the integral, $\int_{4}^{10}{\dfrac{\left[ {{x}^{2}} \right]}{\left[ {{x}^{2}}-28x+196 \right]+\left[ {{x}^{2}} \right]}dx}$.
Let us assume this value as I.
$\Rightarrow I=\int_{4}^{10}{\dfrac{\left[ {{x}^{2}} \right]}{\left[ {{x}^{2}}-28x+196 \right]+\left[ {{x}^{2}} \right]}dx}$
Now, $\left( {{x}^{2}}-28x+196 \right)$ can be written as: -
$\Rightarrow {{x}^{2}}-28x+196={{x}^{2}}-2\times 14\times x+{{14}^{2}}$
The above expression is of the form, ${{a}^{2}}-2ab+{{b}^{2}}$ whose whole square form is, ${{\left( a-b \right)}^{2}}$.
$\Rightarrow {{x}^{2}}-28x+196={{\left( x-14 \right)}^{2}}$
Therefore, the expression of ‘I’ becomes,
$\Rightarrow I=\int_{4}^{10}{\dfrac{\left[ {{x}^{2}} \right]}{\left[ {{\left( x-14 \right)}^{2}} \right]+\left[ {{x}^{2}} \right]}dx}$
$\Rightarrow I=\int_{4}^{10}{\dfrac{\left[ {{x}^{2}} \right]dx}{\left[ {{\left( 14-x \right)}^{2}} \right]+\left[ {{x}^{2}} \right]}}$ - (i)
We know that: - $I=\int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left( a+b-x \right)dx}$, where ‘a’ and ‘b’ are the lower and upper limit of the integral respectively. f (x) is the function inside the sign of integration.

& \Rightarrow I=\int_{4}^{10}{\dfrac{\left[ {{x}^{2}} \right]}{\left[ {{\left( 14-x \right)}^{2}} \right]+\left[ {{x}^{2}} \right]}} \\\ & \Rightarrow f\left( a+b-x \right)=f\left( 4+10-x \right) \\\ & \Rightarrow f\left( a+b-x \right)=f\left( 14-x \right) \\\ & \Rightarrow f\left( a+b-x \right)=\dfrac{\left[ {{\left( 14-x \right)}^{2}} \right]}{\left[ {{x}^{2}} \right]+\left[ {{\left( 14-x \right)}^{2}} \right]} \\\ \end{aligned}$$ So, the value of I can also be written as: - $$\Rightarrow I=\int_{4}^{10}{\dfrac{\left[ {{\left( 14-x \right)}^{2}} \right]dx}{\left[ {{x}^{2}} \right]+\left[ {{\left( 14-x \right)}^{2}} \right]}}$$ - (ii) Adding equations (i) and (ii), we get, $$\Rightarrow 2I=\int_{4}^{10}{\dfrac{\left[ {{x}^{2}} \right]dx}{\left[ {{x}^{2}} \right]+\left[ {{\left( 14-x \right)}^{2}} \right]}}+\int_{4}^{10}{\dfrac{\left[ {{\left( 14-x \right)}^{2}} \right]dx}{\left[ {{x}^{2}} \right]+\left[ {{\left( 14-x \right)}^{2}} \right]}}$$ Taking L.C.M in the right hand side, we get, $$\Rightarrow 2I=\int_{4}^{10}{\dfrac{\left[ {{x}^{2}} \right]+\left[ {{\left( 14-x \right)}^{2}} \right]}{\left[ {{x}^{2}} \right]+\left[ {{\left( 14-x \right)}^{2}} \right]}dx}$$ Cancelling the like terms, we get, $$\begin{aligned} & \Rightarrow 2I=\int_{4}^{10}{1dx} \\\ & \Rightarrow 2I=\int_{4}^{10}{dx} \\\ & \Rightarrow 2I=\left[ x \right]_{4}^{10} \\\ \end{aligned}$$ $$\Rightarrow 2I=\left( 10-4 \right)$$ $$\begin{aligned} & \Rightarrow 2I=6 \\\ & \Rightarrow I=3 \\\ \end{aligned}$$ **So, the correct answer is “Option (c)”.** **Note:** One may note that the property used above to solve the question is one of the most important properties of definite integral that should be remembered. Without using the above property we would have barely been able to solve the question is such an easy manner. That is why we must remember each and every property of definite integral. Sometimes we need to apply more than one property.