## anonymous 5 years ago Evaluate the definite integral. 6 to 4 (3x2−4x+5)dx hi guys i need help ?

1. anonymous

do you know how to integrate polynomials?

2. anonymous

i tried but i could not solve this problem

3. anonymous

Possible intermediate steps: integral (5-4 x+3 x^2) dx Integrate the sum term by term and factor out constants: = 3 integral x^2 dx+ integral 5 dx-4 integral x dx The integral of x^2 is x^3/3: = x^3+ integral 5 dx-4 integral x dx The integral of x is x^2/2: = x^3-2 x^2+ integral 5 dx The integral of 5 is 5 x: = x^3-2 x^2+5 x+constant

4. anonymous

then by applying newton's Newton leibniz you would plug in 4 and 6 to solve.

5. anonymous

=(x^3 - 2x^2 +5x) then evaluate it at 6 then subtract (x^3 - 2x^2 +5x) evaluated at 5

6. anonymous

its a definite integral so no plus c stuff

7. anonymous

Take the indefinite integrate of each individual term using the laws of integration. Next, evalute your answer at each endpoint..then subtract your evaluation at 'a' from your evaluation at 'b'

8. anonymous

$\int\limits_{4}^{6}3x ^{2}-4x +5 dx = [x ^{3}-2x ^{2}+5x ] |_{4}^{6}$

9. anonymous

sorry i meant evaluated at 4 typo

10. anonymous

$\int\limits_{4}^{6} (3x^2-4x+5)dx$

11. anonymous

$6^3 -4(6) + 5(6) -(4^3 -4(4) +5(4))$

12. anonymous

smart, are you still confused? it is pretty straight forward from here. nosmada has explained it just now too.

13. anonymous

$6^3−4(6)^2+5(6)−(4^3−4(4)^2+5(4))$ sorry disregard that last one

14. anonymous

it says the answer is incorrect

15. anonymous

the answer is 82 or not?

16. anonymous

Thanks man for helping me.. it is incorrect.

17. anonymous

no wiat you have to evaluate the polynomial I posted. the one nmosmada posted is incorrect. use mine and solve.

18. anonymous

$6^{3}-2(6)^{2}+5(6)-(4^{3}-2(4)^{2}+5(4))$ that is the correct one. you should get 132 i think

19. anonymous

it's not correct

20. anonymous

oh well, too bad. I did the procedure correctly. maybe your book has an error in it.

21. anonymous

thanks man

22. anonymous

you are welcome. :)