Question

find f(x) when f(-2) = 10 for the anti-derivative (it like a squigly thing ) of
(8x^3 -2x) dx

Answer 1

So this you mean?

\[\Large \int (8x^3-2x)dx\]

this is also called the "integral"

Answer 2

yes

Answer 3

the integral is really the opposite of the derivative

for example

if y = x^2, then dy/dx = 2x

if I told you that dy/dx = 2x and you didn't know that y = x^2, then you'd just think backwards to go from dy/dx = 2x to y = x^2.

BUT

y = x^2 and any other function of the form y = x^2 + C will have the derivative of dy/dx = 2x. Examples: y = x^2 + 1, y = x^2 + 5, etc all have derivatives dy/dx = 2x

so that's why integrals are also referred to as "antiderivatives". They undo derivatives.

Answer 4

can you think of something that may derive to 8x^3 ?

ie

if I told you that dy/dx = 8x^3, then what could y be?

Answer 5

2x^4 ?

Answer 6

very good

you can use a guess and check method, or you can use this rule here

\[\Large \int (x^n)dx = \frac{x^{n+1}}{n+1}+C\]

Notice how deriving the right side will get you to x^n

Answer 7

also, a lot of derivative rules also apply to integrals as well

for instance, with differentiation,we can pull out the coefficient and ignore it for the time being while we derive. The same is true with integration

so what I'm saying is that we can do this

\[\Large \int(8x^3)dx = 8*\int(x^3)dx\]

and then use the rule above

Answer 8

ok

Answer 9

so far I am understanding

Answer 10

great

Answer 11

Using that rule and other integral rules, we get this

\[\Large \int (8x^3-2x)dx = \int(8x^3) + \int(-2x)dx\]

\[\Large \int (8x^3-2x)dx = 8\int(x^3) - 2\int(x)dx\]

\[\Large \int (8x^3-2x)dx = 8\int(x^3) - 2\int(x^1)dx\]

\[\Large \int (8x^3-2x)dx = 8(\frac{x^{3+1}}{3+1}+C) - 2(\frac{x^{1+1}}{1+1}+D)\]

\[\Large \int (8x^3-2x)dx = 8*\frac{x^{4}}{4}+8C - 2*\frac{x^{2}}{2}-2D\]

\[\Large \int (8x^3-2x)dx = 2x^4 - x^2 + C\]

Note: the C and D constants combine to form one constant, and we usually just call it C at the very end

Answer 12

yeah you always add that because you do not know if there is a constant or not

Answer 13

To check the answer, you would derive 2x^4 - x^2 + C and you will get 8x^3 - 2x. So that confirms the antiderivative

Answer 14

I'm assuming your book made F(x) the original function and f(x) the antiderivative

ie, y = f(x), so dy/dx = F(x)

Answer 15

that means

\[\Large \int F(x)dx = f(x) + C\]

Answer 16

but what do I do with the f(-2) = 10 part ? because that was the part that confused me

Answer 17

so essentially we know that

f(x) = 2x^4 - x^2 + C

we also know that f(-2) = 10, so plug in x = -2

f(x) = 2x^4 - x^2 + C

f(-2) = 2(-2)^4 - (-2)^2 + C

then replace f(-2) with 10, since the two are equal, to get

f(-2) = 2(-2)^4 - (-2)^2 + C

10 = 2(-2)^4 - (-2)^2 + C

now solve for C and tell me what you get

Answer 18

ok give me a moment

Answer 19

-18

Answer 20

so this particular f(x) function is really

\[\Large f(x) = 2x^4 - x^2 - 18\]

Answer 21

oh wow , that was so simple thank you so much for your help

Answer 22

you're welcome