Evaluate the definite integral of the algebraic
function. I checked using my calculator and got 38 instead of 34, which, based on my work, is only the result of plugging in the upper limit to the equation without subtracting the result of plugging in the lower limit.

Question

Evaluate the definite integral of the algebraic
function. I checked using my calculator and got 38 instead of 34, which, based on my work, is only the result of plugging in the upper limit to the equation without subtracting the result of plugging in the lower limit.

zzr0ck3r · Answer

function?

anonymous · Answer

$$\int\limits_{1}^{3}(3x ^{2}+5x-4)dx$$

zzr0ck3r · Answer

ok so the integral will be $x^3+\frac{5}{2}x^2-4x|_1^3$ correct?

zzr0ck3r · Answer

@christina18 ?

anonymous · Answer

attachment

zzr0ck3r · Answer

can you answer my question?

anonymous · Answer

That's not how I learned how to evaluate, so I'm not sure.

zzr0ck3r · Answer

$\int (3x^2+5x-4)dx = x^3+\frac{5}{2}x^2-4x+c $

so     $\int_1^3 (3x^2+5x-4)dx = 3^3+\frac{5}{2}3^2-4*3-(1^3+\frac{5}{2}1^2-4*1) =\ 3^3+\frac{5}{2}3^2-4*3-1^3-\frac{5}{2}1^2+4*1=38$

zzr0ck3r · Answer

you need to take the integral before you evaluate the bounds

zzr0ck3r · Answer

what you did was $f(3)-f(1)$

and $f(3)-f(1)\ne \int_1^3(3x^3+5x-4)dx$

zzr0ck3r · Answer

you need to learn how to integrate...

anonymous · Answer

Ok thank you. I know how to integrate, I don't know why I skipped it. Thank you again.

zzr0ck3r · Answer

np