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

wherep(x)=3x2-x+4androle="math" localid="1649347279693" A,B=1,5

Short Answer

Expert verified

When, p(x)=3x2-x+4,A,B=1,5, role="math" localid="1649570449613" ABp(x)dx=128and B-A6p(A)+4pA+B2+p(B)=128which verifies the formula.

Step by step solution

01

Step 1. Given Information

We are given the equation p(x)=3x2-x+4and the interval A,B=1,5.

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

02

Step 2. Concept

The theorem states that, " Ifp(x)is a quadratic equation onA,B, thenrole="math" localid="1649570275366" ABp(x)dx=B-A6p(A)+4pA+B2+p(B)

03

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

We know that p(x)=3x2-x+4and A,B=1,5

Then,

role="math" localid="1649431882297" ABp(x)dx=15(3x2-x+4)dx=153x2dx-15xdx+154dx=3x3315-x2215+4x15=3353-13-1252-12+4(5-1)=125-1-12(25-1)+16=124-12+16=128

04

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

The given interval A,B=1,5. Here, A=1,B=5

Also, p(x)=3x2-x+4

Then,

role="math" localid="1649570206207" p(A)=3(1)2-1+4p(A)=3-1+4=6p(B)=3(5)2-5+4p(B)=75-5+4=74p(A+B2)=p1+52p(3)=3(3)2-3+4=27-3+4=28

Therefore,

B-A6p(A)+4pA+B2+p(B)=5-166+4(28)+74=466+112+74=128

05

Step 5. Conclusion

From the step 3 and 4, we have seen that the formulaABp(x)dx=B-A6p(A)+4pA+B2+p(B)holds forp(x)=3x2-2+4,A,B=1,5

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