91影视

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.

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

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

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

Therefore,

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

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