Question

How do you calculate the derivative of $\int{\sqrt{\left( {{x}^{3}}-2x+6 \right)}}dx$ from $\left[ {{x}^{2}},-2 \right]$?

Answer 1

This type of question is based on the concept of differentiation and integration. Let us consider $\sqrt{\left( {{x}^{3}}-2x+6 \right)}$ to be f(x) and g(x) be ${{x}^{2}}$. We know that $\dfrac{d}{dx}\int\limits_{a}^{x}{f\left( y \right)dy=f\left( x \right)}$ where y is another variable and a is a constant. Here a is -2 and the variable is ${{x}^{2}}$. Therefore, we get $\dfrac{d}{dx}\int\limits_{{{x}^{2}}}^{-2}{f\left( x \right)dx}$. Using the property of definite integral $\int\limits_{a}^{b}{f\left( x \right)}dx=-\int\limits_{b}^{a}{f\left( x \right)dx}$, we can exchange the limits. Use the chain rule of differentiation and do necessary calculations to find the value of $\dfrac{d}{dx}\int\limits_{{{x}^{2}}}^{-2}{\sqrt{\left( {{x}^{3}}-2x+6 \right)}dx}$ which is the required answer.

Complete step by step solution:
According to the question, we are asked to find the derivative of $\int{\sqrt{\left( {{x}^{3}}-2x+6 \right)}}dx$ from $\left[ {{x}^{2}},-2 \right]$.
We have been given the function $\sqrt{\left( {{x}^{3}}-2x+6 \right)}$.
Let us consider the function to be $f\left( x \right)=\sqrt{\left( {{x}^{3}}-2x+6 \right)}$. --------(1)
We know that the derivative of the integral of a function in the limits [a,x] is equal to the function with variable x.
That is $\dfrac{d}{dx}\int\limits_{a}^{x}{f\left( y \right)dy=f\left( x \right)}$.
From the given conditions, we get that the limit is $\left[ {{x}^{2}},-2 \right]$.
The constant a=-2.
Therefore, we get
$\dfrac{d}{dx}\int\limits_{{{x}^{2}}}^{-2}{f\left( x \right)dx}$
But the constant should be at the lower limit.
Using the property of definite integral that is $\int\limits_{a}^{b}{f\left( x \right)}dx=-\int\limits_{b}^{a}{f\left( x \right)}dx$ in the obtained expression, we get
$\dfrac{d}{dx}\int\limits_{{{x}^{2}}}^{-2}{f\left( x \right)dx}=-\dfrac{d}{dx}\int\limits_{-2}^{{{x}^{2}}}{f\left( x \right)dx}$
From the limits, we get that the variable is ${{x}^{2}}$ and not x.
Consider $g\left( x \right)={{x}^{2}}$.
Therefore, using the chain rule, we get
$\dfrac{d}{dx}\int\limits_{{{x}^{2}}}^{-2}{f\left( x \right)dx}=-f\left( g\left( x \right) \right){g}'\left( x \right)$ -----------(2)
Let us find the differentiation of g(x).
$\Rightarrow \dfrac{d}{dx}\left( g\left( x \right) \right)=\dfrac{d}{dx}\left( {{x}^{2}} \right)$
Let us use the power rule of differentiation $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ to solve this.
Here, n=2.
${g}'\left( x \right)=2{{x}^{2-1}}$
On further simplification, we get
${g}'\left( x \right)=2x$
Now, we have to find f(g(x)), that is $f\left( {{x}^{2}} \right)$.
$\Rightarrow f\left( {{x}^{2}} \right)=\sqrt{\left( {{\left( {{x}^{2}} \right)}^{3}}-2{{x}^{2}}+6 \right)}$
On further simplification, we get
$f\left( {{x}^{2}} \right)=\sqrt{\left( {{x}^{6}}-2{{x}^{2}}+6 \right)}$
On substituting in the expression (2), we get
$\dfrac{d}{dx}\int\limits_{{{x}^{2}}}^{-2}{f\left( x \right)dx}=-\sqrt{\left( {{x}^{6}}-2{{x}^{2}}+6 \right)}\left( 2x \right)$
On rearranging the expression, we get
$\dfrac{d}{dx}\int\limits_{{{x}^{2}}}^{-2}{f\left( x \right)dx}=-2x\sqrt{\left( {{x}^{6}}-2{{x}^{2}}+6 \right)}$
Therefore, we get $\dfrac{d}{dx}\int\limits_{{{x}^{2}}}^{-2}{\sqrt{\left( {{x}^{3}}-2x+6 \right)}dx}=-2x\sqrt{\left( {{x}^{6}}-2{{x}^{2}}+6 \right)}$.
Hence, the the derivative of $\int{\sqrt{\left( {{x}^{3}}-2x+6 \right)}}dx$ from $\left[ {{x}^{2}},-2 \right]$ is $-2x\sqrt{\left( {{x}^{6}}-2{{x}^{2}}+6 \right)}$.

Note: We should know the properties of definite integral to solve this type of problems. We should differentiate g(x) also without which the answer is incomplete. Avoid calculation mistakes based on sign convention.