Question

How do you use the second fundamental theorem of Calculus to find the derivative of given \[\int{{{\left( 2t-1 \right)}^{3}}dt}\] from \[\left[ {{x}^{2}},{{x}^{7}} \right]\]?

Answer 1

Hint: This question is from the topic of calculus. In this question, we will find the derivative of given integration. In solving this question, we will first understand about the second fundamental theorem of calculus. After that, we will apply the second fundamental theorem of calculus in solving this question. After applying that theorem, we will solve the further question and get our answer.

Complete step-by-step answer:
Let us solve this question.
In this question, we have to find the derivative of integration given in the question from \[\left[ {{x}^{2}},{{x}^{7}} \right]\] using the second fundamental theorem of calculus. The integration which we have to find the derivative of is:
\[\int{{{\left( 2t-1 \right)}^{3}}dt}\]
For solving the further question, let us first know about second fundamental theorem of calculus.
The second fundamental theorem of calculus says that if a function \[F\left( x \right)\] is given as \[F\left( x \right)=\int{f\left( t \right)dt}\] and the integration limits are given as \[\left[ p\left( x \right),q\left( x \right) \right]\], then we can write the derivative of \[F\left( x \right)\] as \[\dfrac{d}{dx}\left( F\left( x \right) \right)=F'\left( x \right)=f\left( q\left( x \right) \right)\cdot q'\left( x \right)-f\left( p\left( x \right) \right)\cdot p'\left( x \right)\]
Now, let us apply the above theorem for finding the derivative of \[\int{{{\left( 2t-1 \right)}^{3}}dt}\] which limits are given as \[\left[ {{x}^{2}},{{x}^{7}} \right]\]. So, we can write
\[\dfrac{d}{dx}\left( \int\limits_{{{x}^{2}}}^{{{x}^{7}}}{{{\left( 2t-1 \right)}^{3}}dt} \right)\]
We can see that we have written the limits as \[\left[ {{x}^{2}},{{x}^{7}} \right]\]
Using the second fundamental theorem of calculus, we can write the term \[\dfrac{d}{dx}\left( \int\limits_{{{x}^{2}}}^{{{x}^{7}}}{{{\left( 2t-1 \right)}^{3}}dt} \right)\] as
\[\dfrac{d}{dx}\left( \int\limits_{{{x}^{2}}}^{{{x}^{7}}}{{{\left( 2t-1 \right)}^{3}}dt} \right)={{\left( 2\left( {{x}^{7}} \right)-1 \right)}^{3}}\dfrac{d}{dx}\left( {{x}^{7}} \right)-{{\left( 2\left( {{x}^{2}} \right)-1 \right)}^{3}}\dfrac{d}{dx}\left( {{x}^{2}} \right)\]
We can write the above as
\[\Rightarrow \dfrac{d}{dx}\left( \int\limits_{{{x}^{2}}}^{{{x}^{7}}}{{{\left( 2t-1 \right)}^{3}}dt} \right)={{\left( 2{{x}^{7}}-1 \right)}^{3}}\dfrac{d}{dx}\left( {{x}^{7}} \right)-{{\left( 2{{x}^{2}}-1 \right)}^{3}}\dfrac{d}{dx}\left( {{x}^{2}} \right)\]
Using the formula of differentiation: \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\], we can write the above equation as
\[\Rightarrow \dfrac{d}{dx}\left( \int\limits_{{{x}^{2}}}^{{{x}^{7}}}{{{\left( 2t-1 \right)}^{3}}dt} \right)={{\left( 2{{x}^{7}}-1 \right)}^{3}}\left( 7{{x}^{6}} \right)-{{\left( 2{{x}^{2}}-1 \right)}^{3}}\left( 2x \right)\]
So, we have solved this question now. And, we have found the derivative of \[\int\limits_{{{x}^{2}}}^{{{x}^{7}}}{{{\left( 2t-1 \right)}^{3}}dt}\]. The derivative is: \[{{\left( 2{{x}^{7}}-1 \right)}^{3}}\left( 7{{x}^{6}} \right)-{{\left( 2{{x}^{2}}-1 \right)}^{3}}\left( 2x \right)\]

Note: We should have a better knowledge in the topic of calculus to solve this type of question easily. We should remember the following formula of differentiation:
\[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]
Remember the second fundamental theorem of calculus. This theorem says that if \[F\left( x \right)=\int\limits_{p\left( x \right)}^{q\left( x \right)}{f\left( t \right)dt}\], then its derivative \[F'\left( x \right)=\dfrac{d}{dx}\left( \int\limits_{p\left( x \right)}^{q\left( x \right)}{f\left( t \right)dt} \right)\] can be written as
\[F'\left( x \right)=f\left( q\left( x \right) \right)\cdot q'\left( x \right)-f\left( p\left( x \right) \right)\cdot p'\left( x \right)\]
Remember that \[F\left( x \right)\] should be continuous from \[p\left( x \right)\] to \[q\left( x \right)\].