Question

int_{-1}^{3}(2-x-5x)dx
Evaluate the above integral using the Fundamental Theorem of Calculus.

Answer 1

`Fundamental Theorem of Calculus, Part 2:`
\[\Large \int\limits_a^b f(x)\;dx \quad=\quad F(b)-F(a)\]Where `F` is the anti-derivative of `f`.
Hmm ok do you understand how to apply the power rule to each term within the integral? :o

Answer 2

i did it and i got (-131/3) is it correct?

Answer 3

Oh lemme check a sec :3

Answer 4

okay

Answer 5

Hmm wolfram is saying -16, is this what the function looks like?\[\Large \int\limits_{-1}^3 2-x-5x\;dx\]

Answer 6

yes

Answer 7

\[2(x)-\frac{ x^2 }{ 2 }-\frac{ 5(x^3) }{ 3 } \]
thats waht i did

Answer 8

Was the 5x supposed to be 5x^2?

Answer 9

yes \[\int\limits_{-1}^{3}(2x-x-5x^2)\]

Answer 10

oh..

Answer 11

yep

Answer 12

Hmm seems like it should work out to -128/3.
Maybe you just made a minor mistake near the end somewhere.

Answer 13

okay ill try to do it again. let you know when i done.

Answer 14

k imma go make a sammich c:

Answer 15

\[\int\limits_{-1}^{3} 2-x-5x^2\]
\[2(x)-\frac{ x^2 }{ 2 }-\frac{ 5(x^3) }{ 3 }\]
\[[2(3)-\frac{ 3^2 }{ 2 }-\frac{ 5(3^3) }{ 3 }]-[2(-1)-\frac{ -1^2 }{ 2 }-\frac{ 5(-1^3) }{ 3 }]\]
\[(-43.5)+(-0.1666666667) =\frac{ -131 }{ 3 }\]

Answer 16

thats how i did it and i got the same answer

Answer 17

\[\Large \left[6-\frac{9}{2}-45\right]-\left[-2\color{red}{-\frac{1}{2}}+\frac{5}{3}\right]\]This red part, did you remember to square the negative sign?

Answer 18

\[\Large \left[6-\frac{9}{2}-45\right]-\left[-2\color{red}{-\frac{(-1)^2}{2}}+\frac{5}{3}\right]\]

Answer 19

oh gotcha!! thanks for your time i really appreciate