Q. 13 - Calculus

Question

Calculate each definite integral in using

Part (a): The definition of the definite integral as a limit of Riemann sums.

Part (b): The definite integral formulas from Theorem 4.13.

Part (c): the Fundamental Theorem of Calculus. Then show that your three answers are the same.

${\int }_1^{2}3x^2 dx$

Step-by-Step Solution

Answer

Part (a): $\underset{n\to \infty }{\lim }\left[\frac{3}{n}\left(n\right)+\frac{6}{n^2}\left({\displaystyle \frac{n\left(n+1\right)}{2}}\right)+\frac{3}{n^3}\left({\displaystyle \frac{n\left(n+1\right)\left(2n+1\right)}{2}}\right)\right]=7$

Part (b): $3\left(\frac{1}{3}\right)\left(2^3-1^3\right)=7$

Part (c): ${{\left[3\left(\frac{1}{3}\right)x^3\right]}^2}_1=7$

Part (a) Step 1. Given information.

Consider the given question,

${\int }_1^{2}3x^2 dx$

Part (a) Step 2. Using limit of Riemann sums.

The right sum defined for n rectangles on $\left[a,b\right]$ is $\sum _{k=1}^{n}f\left(x_k\right)△x$.

Where, $$\begin{gathered} △x=\frac{b-a}{n}, \\ {x}_k=a+k△x \end{gathered}$$

The interval is $\left[1,2\right]$. Now $△x$,

$$\begin{gathered} =\frac{2-1}{n} \\ =\frac{1}{n} \end{gathered}$$

Then $x_k$,

$$\begin{gathered} =1+k\left(\frac{1}{n}\right) \\ =1+\frac{k}{n} \end{gathered}$$

Part (a) Step 3. Write the right sum.

Consider the right sum,

$$\begin{gathered} =\sum _{k=1}^n\left(3{\left(1+\frac{k}{n}\right)}^2\right)\left(\frac{1}{n}\right) \\ =\frac{3}{n}\sum _{k=1}^{n}1+\frac{2k}{n}+\frac{k^2}{n^2} \\ =\frac{3}{n}\sum _{k=1}^{n}1+\frac{6}{n^2}\sum _{k=1}^{n}k+\frac{3}{n^3}\sum _{k=1}^n{k}^2 \\ =\frac{3}{n}\left(n\right)+\frac{6}{n^2}\left({\displaystyle \frac{n\left(n+1\right)}{2}}\right)+\frac{3}{n^3}\left({\displaystyle \frac{n\left(n+1\right)\left(2n+1\right)}{2}}\right) \end{gathered}$$

Then,

$$\begin{gathered} {\int }_0^{1}3x^2 dx=\underset{n\to \infty }{\lim }\left[\frac{3}{n}\left(n\right)+\frac{6}{n^2}\left({\displaystyle \frac{n\left(n+1\right)}{2}}\right)+\frac{3}{n^3}\left({\displaystyle \frac{n\left(n+1\right)\left(2n+1\right)}{2}}\right)\right] \\ =7 \end{gathered}$$

Therefore, the value is $7$.

Part (b) Step 1. Using definite integral formula.

The value is given below,

$$\begin{gathered} ={\int }_0^{1}3x^2 dx \\ =3\left(\frac{1}{3}\right)\left(2^3-1^3\right) \\ =8-1 \\ =7 \end{gathered}$$

Therefore, the value is $7$.

Part (c) Step 1. Using the fundamental theorem.

The value is given below,

$$\begin{gathered} ={\int }_0^{1}3x^2 dx \\ ={{\left[3\left(\frac{1}{3}\right)x^3\right]}^2}_1 \\ =2^3-1^3 \\ =8-1 \\ =7 \end{gathered}$$