# Math Help - Integral word problem

1. ## Integral word problem

I have attempted to answer this question but come up with a number that just doesn't seem correct....please review my solution and let me know where I went wrong. Thanks!

Find f(2) if $f(1/3) = 2$ and $f ' (x) = e^{3x+2}-3x$

my solution:
= $\frac{1}{3}e^{3x+2}+C$

and since f(1/3) = 2 then: 2 = $\frac{1}{3}e^{3(1/3)+2}+C$
and solving for C gives me a not so great number.

please let me know where I went wrong....thanks!

I have attempted to answer this question but come up with a number that just doesn't seem correct....please review my solution and let me know where I went wrong. Thanks!

Find f(2) if $f(1/3) = 2$ and $f ' (x) = e^{3x+2}-3x$

my solution:
= $\frac{1}{3}e^{3x+2}+C$

and since f(1/3) = 2 then: 2 = $\frac{1}{3}e^{3(1/3)+2}+C$
and solving for C gives me a not so great number.

please let me know where I went wrong....thanks!

$f'(x)=e^{3x+2}-3x\ \Rightarrow\ f(x)=\frac{1}{3}e^{3x+2}-\frac{3}{2}x^2+C$

Now, C is found from $f\left(\frac{1}{3}\right)=2$

Having found C, f(2) can be evaluated.

3. I understand how you got your answer but the answers I get solving for C and 2 still don't seem right....but I guess working with 'e' is usually like that.

Thank you

4. You just need to practice lil_cookie,

$f\left(\frac{1}{3}\right)=e^{\frac{3}{3}+2}-\frac{3}{2}\left(\frac{1}{3}\right)^2+C=e^3-\frac{1}{6}+C=2$

$C=\frac{13}{6}-e^3$

Then

$f(2)=e^{6+2}-\frac{3}{2}2^2+\frac{13}{6}-e^3=e^8-6+\frac{13}{6}-e^3$

5. oh you don't actually solve for e

thank you

The final numerical value of f(2) is $e^8-e^3-\frac{36}{6}+\frac{13}{6}$
$=2980.957987-20.085537-\frac{23}{6}=2957.04.$