Question

How will you evaluate the definite integral \[\int{3{{x}^{2}}dx}\] from [1, 2]?

Answer 1

Hint: We are asked to find the integral of \[3{{x}^{2}}\] from [1, 2]. We will learn about the integral and we will learn which term can be integrated and which cannot and which terms are taken out of integration. After that, we will use \[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}\] to solve the problem. In our problem, we are also given the limit of integration, so we will use that too.

Complete step by step answer:
We are given \[\int{3{{x}^{2}}dx}\] on [1, 2]. Before we start, we should know that integration is the method in which we combine the small quantity and solve for the large. In integration, there is no effect on the constant, we can take it outside our integration that is \[\int{k}dx=k\int{dx}\] where k is some constant. We know that the formula of integration is \[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}.\] In our integral, we have the domain where we need to integrate, so it means we have the limit of the integral.
Once, we have the value of our integral, then we apply this limit. We put x = 2 and x= 1 and subtract the value. As we can see that, in \[\int{3{{x}^{2}}dx}\] 3 is constant. So, we take it outside. Now, we have \[3\int\limits_{1}^{2}{{{x}^{2}}}dx\] as we can see that we have n = 2. Using the formula \[\int\limits_{a}^{b}{{{x}^{n}}}dx=\left[ \dfrac{{{x}^{n+1}}}{n+1} \right]_{a}^{b}\] we will get
\[3\int\limits_{1}^{2}{{{x}^{2}}dx}=3\left[ \dfrac{{{x}^{2+1}}}{2+1} \right]_{1}^{2}\]
On simplifying, we get,
\[\Rightarrow 3\int\limits_{1}^{2}{{{x}^{2}}dx}=3\times \left[ \dfrac{{{x}^{3}}}{3} \right]_{1}^{2}\]
Cancelling the like terms, we get,
\[\Rightarrow 3\int\limits_{1}^{2}{{{x}^{2}}dx}=\left[ {{x}^{3}} \right]_{1}^{2}\]
Putting the limit, we get,
\[\Rightarrow 3\int\limits_{1}^{2}{{{x}^{2}}dx}={{\left( 2 \right)}^{3}}-{{\left( 1 \right)}^{3}}\]
As, \[{{2}^{3}}=8,\] and \[{{1}^{3}}=1,\] so we get,
\[\Rightarrow 3\int\limits_{1}^{2}{{{x}^{2}}dx}=8-1\]
So, we get,
\[\Rightarrow 3\int\limits_{1}^{2}{{{x}^{2}}dx}=7\].

Note:
Remember that in the case of infinite integral, that is integral without the limit, it is necessary to replace or substitute back the original value of the function back which we replaced earlier. While in the case of definite integral, no such thing is necessary as we change the function, we also change the limit along with it. So, we did not require to change back to the function. We simply apply the limit there but yes with the change in the function, the limit must also be changed accordingly.