Question

Can someone please help me with Calculus 1 problems ? :)

Answer 1

What's the problem?

Answer 2

its \[f ' (x) = 2x ^{2/3}+4x ^{5/3}-3x+\sqrt{5}\]

Answer 3

Find f (x) when f ' (x) is given

Answer 4

You have to do that by integration, because differentiation is
f(x)-->f'(x)
And so integration will be
f'(x)-->f(x)
\[\int\limits( 2x ^{2/3}+4x ^{5/3}-3x+\sqrt{5})dx\]

Answer 5

You can apply power rule for each term, can you do that?

Answer 6

can you please guide me little bit more?

Answer 7

There are 4 terms here, each can be used by power rule separately, by applying the power rule formula which is:
\[\large \int\limits x^n~dx=\frac{x^{n+1}}{n+1}\]
------------------------------------------------
You get:
\[\int\limits \left(2 x^{2/3}+4 x^{5/3}-3 x+\sqrt{5}\right) \, dx=\frac{6 x^{5/3}}{5}+\frac{3 x^{8/3}}{2}-\frac{3 x^2}{2}+\sqrt{5} x\]
Do you agree?

Answer 8

I m trying to understand that How you got 6x^5/3/5+3x^8/3/2-3x^2/2+radical of 5x

Answer 9

if you dont mind just to make my life easier, can you please just show me little bit detail work for 6x^5/3/5 then I will try to figure out rest.