Antiderivative
- Thread starter JenniferC
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You have actually done the hard part
\(\displaystyle f'(x) = 2\sqrt{x} + 2 \implies \displaystyle \int f'(x)\ dx = \int 2x^{(1/2)} + 2\ dx = \dfrac{4x^{(3/2)}}{3} + 2x + C.\)
\(\displaystyle So\ f(1) = 3 \implies 3 = \dfrac{4 * 1^{(3/2)}}{3} + 2(1) + C \implies\)
\(\displaystyle 3 = \dfrac{4 * 1}{3} + 2 + C \implies 9 = 4 + 6 + 3C \implies -1 = C \implies C = -\dfrac{1}{3}.\)
So what does f(x) equal? So what is f(0)?
If you are given a single value of f(x), you can solve for C.
Edit: As pka says below, you did not write out the antiderivative correctly.
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pka
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That is incorrect.
\(\displaystyle f(x) = \dfrac{{4\sqrt {{x^3}} }}{3} + 2x + c\)
Then \(\displaystyle 3 = f(1) = \dfrac{{4\sqrt {{1^3}} }}{3} + 2 \cdot 1 + c\) so \(\displaystyle c=~?\)
Please see my edit to my first post and pka's post. But you have the general idea.
Thanks. I left out the "/ 3." You still get the same answer though.
Having left off the / 3, you would not have calculated C correctly from f(1). You have to get the antiderivative right in the first place to compute C correctly. It is this kind of error that kills you on exams. I am quite confident that you understand antiderivatives and that you now understand when and how you can compute C, but, unless you check your mechanics, you will make careless errors. Pka gave you a better answer than I did because he did check your mechanics.
There is a double lesson here for you. The one intended on when and how C can be found. The other is to remember to differentiate your antiderivative to check that you get the derivative you started with.