Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can somebody please show me how to evaluate the following integral: $$\int_0^1 t\sqrt{4+9t^2} \,dt?$$

I know how to integrate a square root function, but the $t$ in front throws me off, and if there are associated properties with this sort of integral.

Could you also, knowing that $t=\sqrt{t^2}$, multiply it into the other square root and integrate that way?

share|cite|improve this question
for a less algorithmic way of solving this problem see… – John Joy Sep 2 '14 at 13:03

Hint: Substitute $u = 4 + 9t^2$, $du = 18 t dt$.

share|cite|improve this answer
o jeez thanks... – user89853 Aug 11 '13 at 0:12

Let $\;\color{red}{\bf u = 4 + 9t^2}.\;$ Then $ \,du = 18 t \,dt \iff \color{blue}{\bf t\,dt = \dfrac 1{18}} \,du$.

Now, for our new bounds of integration: when $\;t = 0, \;u = 4 + 9(0)^2 = 9.\;$ When $\;t = 1, \;u = 4+9(1)^2 = 13$.

Now substitute, to get $$\int_0^1 \sqrt{\color{red}{\bf 4 + 9t^2}}\color{blue}{\bf (t\,dt)} \quad = \quad \color{blue}{\bf \dfrac1{18}}\int_{\bf 4}^{\bf 13} \sqrt{\color{red}{\bf u}} \color{blue}{\bf \,du}\quad = \quad \dfrac 1{18} \int_4^{13} u^{1/2} \,du$$

Integrate, and then evaluate at the new bounds of integration, so there's no need to back-substitute.

share|cite|improve this answer


$$\int f'(x)f^n(x)dx=\frac{f^{n+1}(x)}{n+1}$$

but $f(x)=4+9t^2$ so $f'(x)=18t$

so $$\int t\sqrt{4+9t^2}dx=\frac{1}{18}\int 18t\sqrt{4+9t^2}dx=\frac{1}{27}(\sqrt{(4+9t^2)^3})+C$$

where C is arbitrary constant

share|cite|improve this answer

You say:

I know how to integrate a square root function

But more exactly, you know how to integrate $$\int u^\frac{1}{2} du$$ So, if you make the thing under the square root sign u, does the rest of the integrand become $du$? This is the sort of question that leads, in this case, to the substitution of the responders above...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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