Question

Use the Fundamental Theorem to evaluate (Your answer must include the antiderivative.) –

Answer 1

@SolomonZelman

Answer 2

@jabez177

Answer 3

FTofC \[\int_b^a f(x) dx = F(b) - F(a)\]

Answer 4

There are certain rules of integration that you have to apply:

At first, you need to know that:
\(\color{#000000 }{ \displaystyle [1]\quad \int_a^b\left(f(x)+g(x)\right){\tiny~}dx= \int_a^bf(x){\tiny~}dx+\int_a^bg(x){\tiny~}dx }\)\(\tiny\\[1.9em]\)
\(\color{#000000 }{ \displaystyle [2]\quad \int_a^b\left(f(x)-g(x)\right){\tiny~}dx= \int_a^bf(x){\tiny~}dx-\int_a^bg(x){\tiny~}dx }\)

(These are true by indefinite integrals as well)

Answer 5

Yep, integrals are linear. That will help too.

Answer 6

And we need to know the power rule,

\(\color{#000000 }{ \displaystyle [3]\quad \int x^n~dx=\frac{x^{n+1}}{n+1}+C }\)

Answer 7

Okay well I think before I Use the Fundamental Theorem to evaluate I have to find the antiderivative.

Answer 8

I can give you an example, if you like ...

Answer 9

\(\color{#000000 }{ \displaystyle {\rm Example:}\quad\quad \int_2^3 \left(x^4+8\right){\tiny~}dx}\)

I know that "the integral of a sum, is a sum of integrals". (rule [1])
\(\color{#000000 }{ \displaystyle \int_2^3 x^4 {\tiny~}dx+\int_2^3 8 {\tiny~}dx}\)

Then, I apply the power rule, term by term:
\(\color{#000000 }{ \displaystyle \int_2^3 x^4 {\tiny~}dx=\left.\frac{x^{4+1}}{4+1}\right|_{x~=2}^{x~=3}}\)

\(\color{#000000 }{ \displaystyle \int_2^3 8 {\tiny~}dx=\int_2^3 8x^0 {\tiny~}dx=8\times \left.\frac{x^{0+1}}{0+1}\right|_{x~=2}^{x~=3}}\)

So, overall my integrals becomes:
\(\color{#000000 }{ \displaystyle \int_2^3 x^4 {\tiny~}dx=\left.\frac{x^{5}}{5}\right|_{x~=2}^{x~=3}=\left[\frac{(3)^5}{5}\right]-\left[\frac{(2)^5}{5}\right]}\)


\(\color{#000000 }{ \displaystyle\left. \int_2^3 8x^0 {\tiny~}dx=x\right|_{x~=2}^{x~=3}=\left[(3)\right]-\left[(2)\right]}\)

Answer 10

You can write it differently, but, with the same idea...

Answer 11

\(\color{#000000 }{ \displaystyle \left.\int_2^3\left(x^4+8\right)~dx=\left[\frac{x^{4+1}}{4+1}+8x\right]\right|_{x~=2}^{x~=3} }\)

\(\color{#000000 }{ \displaystyle\int_2^3\left(x^4+8\right)~dx=\left[\frac{\color{red}{3}^{4+1}}{4+1}+8\cdot \color{red}{3}\right]-\left[\frac{\color{red}{2}^{4+1}}{4+1}+8\cdot \color{red}{2}\right] }\)

and whatever that comes down to...

Answer 12

(Note, that I showed why the integral of 8 is 8x, and the integral of any "a" would be ax)

Answer 13

srry im working this out

Answer 14

\(\color{#000000 }{ \displaystyle \left.\int_3^6\left(x^2+3\right)~dx=\left[\frac{x^{2+1}}{2+1}+3x\right]\right|_{x~=3}^{x~=6} \\[1.9em] \displaystyle =\left[\frac{\color{red}{6}^{3}}{3}+3\cdot\color{red}{6}\right]-\left[\frac{\color{red}{3}^{3}}{3}+3\cdot\color{red}{3}\right]=\left[\frac{216}{3}+18\right]-\left[\frac{27}{3}+9\right] \\[1.9em] \displaystyle =\left[72+18\right]-\left[9+9\right] =72+18-18=72 }\)

Answer 15

that is another example ...

Answer 16

Yes that is correct.

Answer 17

(not done, but so far...)

Answer 18

i know

Answer 19

Good, continue ... :)

Answer 20

I think I got it from here :) Thank you

Answer 21

OK:) YW