Question

find an antiderivative F(x) with F'(x)=f(x) and F(0)=7. f(x)=5x^2+12x+3

Answer 1

I believe it is \[5x^{2}+12x+3 \rightarrow \frac{5}{3}x^{3}+\frac{12}{2}x^{2}+3x\]

Answer 2

F(0) has to equal 7. I forgot to add that

Answer 3

Well, if you have F'(x)=\[\frac{5}{3}x^{3}+\frac{12}{2}x^{2}+3x+c\] then F'(0) = \[\frac{5}{3}0^{3}+\frac{12}{2}0^{2}+30+c=7\rightarrow c=7\]

Answer 4

Thank you so much :)

Answer 5

You're welcome.