Question

The value of $\int_0^2 {{x^{\left[ {{x^2} + 1} \right]}}dx}$ where $\left[ x \right]$ is the greatest integer less than or equal to $x$ is:
A.$\dfrac{{469}}{{60}} - \dfrac{1}{3}{2^{3/2}} - \dfrac{1}{5}{3^{5/2}}$
B.$\dfrac{{469}}{{60}} + \dfrac{1}{3}{2^{3/2}} + \dfrac{1}{5}{3^{5/2}}$
C.$\dfrac{{469}}{{60}} + \dfrac{1}{3}{2^{3/2}} - \dfrac{1}{5}{3^{5/2}}$
D.0

Answer 1

The greatest integer function is also known as floor function and it represents the greatest integer less than or equal to $x$ and when applied on any real number, it yields an integer. We can use the definition of this greatest integer function on the power of $x$ as is given in the question and evaluate the result.

Formula Used:
We shall be the following formula to calculate the value of integral:
$\int_n^{n + 3} {xdx}$= $\int_n^{n + 1} {xdx}$+ $\int_{n + 1}^{n + 2} {xdx}$+ $\int_{n + 2}^{n + 3} {xdx}$ …(i)

Complete step-by-step answer:
Now, we shall be using the definition of the greatest integer function to calculate the value of the expression passed as an argument to the greatest integer function within its limits.
In the question, the limit of the integral is from $0$to $2$, but the expression passed as the argument of the greatest integer function has ${x^2}$in it, so, the limit of the greatest integer function is going to be from ${0^2}$=$0$ to ${2^2}$= $4$, $0$to$4$.
Now, applying the greatest integer function to the limits, we have:
In the interval $[0,1)$, $\left[ {{x^2} + 1} \right]$ = $1$
In the interval $[1,\sqrt 2 )$, $\left[ {{x^2} + 1} \right]$ = $2$
And, in the interval $[\sqrt 2 ,\sqrt 3 )$, $\left[ {{x^2} + 1} \right]$ = $3$
Finally, in the interval $[\sqrt 3 ,2)$, $\left[ {{x^2} + 1} \right]$ = $4$
=$\int_0^2 {{x^{\left[ {{x^2} + 1} \right]}}dx}$= $\int_0^1 {{x^{\left[ {{x^2} + 1} \right]}}dx}$+ $\int_1^{\sqrt 2 } {{x^{\left[ {{x^2} + 1} \right]}}dx}$+ $\int_{\sqrt 2 }^{\sqrt 3 } {{x^{\left[ {{x^2} + 1} \right]}}dx}$+ $\int_{\sqrt 3 }^2 {{x^{\left[ {{x^2} + 1} \right]}}dx}$
Now, putting in the values obtained from the interval calculation into the values of the integral, we have:
=$\int_0^2 {{x^{\left[ {{x^2} + 1} \right]}}dx}$= $\int_0^1 {{x^1}dx}$+ $\int_1^{\sqrt 2 } {{x^2}dx}$+ $\int_{\sqrt 2 }^{\sqrt 3 } {{x^3}dx}$+ $\int_{\sqrt 3 }^2 {{x^4}dx}$
Solving the integrals by applying the formula of integration, we have got:
= $\dfrac{1}{2}\left[ {{x^2}} \right]_0^1$+ $\dfrac{1}{3}\left[ {{x^3}} \right]_1^{\sqrt 2 }$+ $\dfrac{1}{4}\left[ {{x^4}} \right]_{\sqrt 2 }^{\sqrt 3 }$+ $\dfrac{1}{5}\left[ {{x^5}} \right]_{\sqrt 3 }^2$
Putting in the upper and the lower limits in the brackets and then solving the equations, we have:
= $\dfrac{1}{2} \times 1$ + $\dfrac{1}{3}\left( {{2^{3/2}} - 1} \right)$+ $\dfrac{1}{4}\left[ {9 - 4} \right]$ + $\dfrac{1}{5}\left[ {32 - {3^{5/2}}} \right]$
Now, simplifying the equation we get:
= $\dfrac{{469}}{{60}} + \dfrac{1}{3}{2^{3/2}} - \dfrac{1}{5}{3^{5/2}}$
Hence, the answer to the given question is $C$ $\dfrac{{469}}{{60}} + \dfrac{1}{3}{2^{3/2}} - \dfrac{1}{5}{3^{5/2}}$.

Note: In solving these types of questions, always try to reduce the complex expressions into smaller, simpler ones. This helps us to solve the question way more easily and it also keeps the writing space tidy. Also, special care must be given to the calculation as a lot of numbers are used and at times, mistakes occur simply because of mess.