I have this math problem:

enter image description here

First: I found the integral from $-1$ to $3$ of $|x^2+3|$ and got $64/3$

Second: I found the integral from $-1$ to $3$ of $|x|$ and got $5$

Finally: I subtracted $5$ from $64/3$ and got $49/3$

However $49/3$ is wrong, what did I do wrong?

  • 1
    $\begingroup$ Did you really mean the absolute values $|\cdots|$ ? $\endgroup$ – Han de Bruijn Sep 28 '14 at 19:12

you have to use the form $$\int_a^b f(x)-g(x)dx$$ since the first integral contains a negative part of the function $y=x$ and this part gives negative area i.e. the value 5 is not true
The right one is $$\int_{-1}^3 x^2+3-xdx=52/3$$

  • $\begingroup$ So, the answer is 52/3? $\endgroup$ – KFC Sep 28 '14 at 19:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.