Question

If ${I_1} = \int\limits_1^2 {x\left[ {\sqrt x + \sqrt {3 - x} } \right]dx}$ and ${I_2} = \int\limits_1^2 {\left[ {\sqrt x + \sqrt {3 - x} } \right]dx}$ then $\dfrac{{{I_1}}}{{{I_2}}} =$

1. $\dfrac{1}{2}$
2. $\dfrac{3}{2}$
3. 2
4. 1

Answer 1

We will first substitute the value of ${I_1} = \int\limits_1^2 {x\left[ {\sqrt x + \sqrt {3 - x} } \right]dx}$ and ${I_2} = \int\limits_1^2 {\left[ {\sqrt x + \sqrt {3 - x} } \right]dx}$ in the given expression $\dfrac{{{I_1}}}{{{I_2}}}$ and then simplify it. We will use the rule of integration to integrate the simplified value and at last apply the limits to get the required answer.

Complete step-by-step answer:
We are given that ${I_1} = \int\limits_1^2 {x\left[ {\sqrt x + \sqrt {3 - x} } \right]dx}$ and ${I_2} = \int\limits_1^2 {\left[ {\sqrt x + \sqrt {3 - x} } \right]dx}$
We have to find the value of $\dfrac{{{I_1}}}{{{I_2}}}$
We will substitute the value of ${I_1}$ and ${I_2}$ in $\dfrac{{{I_1}}}{{{I_2}}}$
Then,
=$\dfrac{{\int\limits_1^2 {x\left[ {\sqrt x + \sqrt {3 - x} } \right]dx} }}{{\int\limits_1^2 {\left[ {\sqrt x + \sqrt {3 - x} } \right]dx} }}$
We will simplify the above expression,
=$\int\limits_1^2 {\dfrac{{x\left[ {\sqrt x + \sqrt {3 - x} } \right]}}{{\left[ {\sqrt x + \sqrt {3 - x} } \right]}}} dx$
We will cancel the common terms
=$\int\limits_1^2 {xdx}$
Now, we know that $\int\limits_a^b {xdx = \left( {\dfrac{{{x^2}}}{2}} \right)} _a^b$
Therefore, the value of the integration is
=$\left( {\dfrac{{{x^2}}}{2}} \right)_1^2$
On substituting the limits in the above function, we will get,
=$\left( {\dfrac{{{2^2}}}{2}} \right) - \left( {\dfrac{{{1^2}}}{2}} \right) = \dfrac{4}{2} - \dfrac{1}{2} = \dfrac{3}{2}$
Hence, the value of $\dfrac{{{I_1}}}{{{I_2}}}$ is $\dfrac{3}{2}$
Thus, option (2) is correct.

Note: Integration is the reverse process of differentiation. We can also do this question by solving each integral ${I_1}$ and ${I_2}$ separately and then substituting its value, but that would make the question lengthy. And do not forget to apply given limits of integration at the end. We have to apply the upper limit first and then subtract the value of integration corresponding to the lower limit from it.