Question

Verify that the formula in Theorem 5.28 is true in the case

where$p\left(x\right)=3x^2-x+4$androle="math" localid="1649347279693" $\left[A,B\right]=\left[1,5\right]$

Short answer

When, $p\left(x\right)=3x^2-x+4,\left[A,B\right]=\left[1,5\right]$, role="math" localid="1649570449613" ${\int }_A^{B}p\left(x\right)dx=128$and $\frac{B-A}{6}\left(p\left(A\right)+4p\left(\frac{A+B}{2}\right)+p\left(B\right)\right)=128$which verifies the formula.

Step-by-step solution

Step 1. Given Information

We are given the equation $p\left(x\right)=3x^2-x+4$and the interval $\left[A,B\right]=\left[1,5\right]$.

We need to verify the theorem 5.28 in the case of these above-mentioned values.

Step 2. Concept

The theorem states that, " If$p\left(x\right)$is a quadratic equation on$\left[A,B\right]$, thenrole="math" localid="1649570275366" ${\int }_A^{B}p\left(x\right)dx=\frac{B-A}{6}\left(p\left(A\right)+4p\left(\frac{A+B}{2}\right)+p\left(B\right)\right)$

Step 3. Finding the value of ∫ABp(x)dx 

We know that $p\left(x\right)=3x^2-x+4$and $\left[A,B\right]=\left[1,5\right]$

Then,

role="math" localid="1649431882297" $$\begin{gathered} {\int }_A^{B}p\left(x\right)dx={\int }_1^5(3x^2-x+4)dx \\ ={\int }_1^{5}3x^{2}dx-{\int }_1^{5}xdx+{\int }_1^{5}4dx \\ ={\left[\frac{3x^3}{3}\right]}_1^5-{\left[\frac{x^2}{2}\right]}_1^5+{\left[4x\right]}_1^5 \\ =\left[\frac{3}{3}\left(5^3-1^3\right)\right]-\left[\frac{1}{2}\left(5^2-1^2\right)\right]+\left[4(5-1)\right] \\ =\left[125-1\right]-\left[\frac{1}{2}(25-1)\right]+16 \\ =124-12+16 \\ =128 \end{gathered}$$

Step 4. Finding the value of B-A6p(A)+4pA+B2+p(B)

The given interval $\left[A,B\right]=\left[1,5\right]$. Here, $A=1,B=5$

Also, $p\left(x\right)=3x^2-x+4$

Then,

role="math" localid="1649570206207" $$\begin{gathered} \mathit{p}\mathbf{\left(}\mathit{A}\mathbf{\right)}=3{\left(1\right)}^2-1+4 \\ \Rightarrow p\left(A\right)=3-1+4=6 \\ \mathit{p}\mathbf{\left(}\mathit{B}\mathbf{\right)}=3{\left(5\right)}^2-5+4 \\ \Rightarrow p\left(B\right)=75-5+4=74 \\ \mathit{p}\left(\frac{A+B}{2}\right)=p\left(\frac{1+5}{2}\right) \\ \Rightarrow p\left(3\right)=3{\left(3\right)}^2-3+4 \\ =27-3+4=28 \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{B-A}{6}\left[p\left(A\right)+4p\left(\frac{A+B}{2}\right)+p\left(B\right)\right]=\frac{5-1}{6}\left[6+4\left(28\right)+74\right] \\ =\frac{4}{6}\left[6+112+74\right] \\ =128 \end{gathered}$$

Step 5. Conclusion

From the step 3 and 4, we have seen that the formula${\int }_A^{B}p\left(x\right)dx=\frac{B-A}{6}\left(p\left(A\right)+4p\left(\frac{A+B}{2}\right)+p\left(B\right)\right)$holds for$p\left(x\right)=3x^2-2+4,\left[A,B\right]=\left[1,5\right]$