Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$ \int_0^4 \frac{\mathrm d x}{\sqrt{2x+1}}$$

Then:

  • $t=2x+1$
  • $\mathrm dt=2 \mathrm dx$
  • $\mathrm dx=\mathrm dt /2$

$$ \int_0^4 \frac{\mathrm d t}{2\sqrt t} $$

And what next?

share|improve this question
    
Write ${1\over 2\sqrt t}={1\over 2} t^{-1/2}$. You also need to change your limits of integration... –  David Mitra Jan 20 '13 at 17:37

3 Answers 3

You have done the "hard" part which is seeing the substitution, now everything you have to do is solve that integral which is really easy if you write it this way: $$1/2\int_a^b t^{-1/2}dt$$ Be carefull with the limits: $$x=0\Rightarrow t=2x+1=1$$ $$x=4\Rightarrow t=2\cdot 4+1=9$$ The integral will be: $$\frac{1}{2}\int_1^9t^{-\frac{1}{2}}dt=\frac{1}{2}\left.\frac{t^\frac{1}{2}}{\frac{1}{2}}\right|_{1}^{9}=9^{1/2}-1^{1/2}=3-1=2$$

share|improve this answer

The integral is straightforward $$\int_0^4 \frac{\mathrm d x}{\sqrt{2x+1}}=\int_0^4 \left(\sqrt{2x+1}\,\right)^\prime \,\mathrm d x= \left(\sqrt{2x+1}\,\right)\bigg|_0^4=3-1=2.$$

share|improve this answer

$$\int_0^4 \frac{dx}{\sqrt{2x+1}} = \frac{1}{2}\int_{1}^{9} \frac{du}{\sqrt{u}} = \frac{1}{2}\int_{1}^{9} u^{-1/2}\, du$$

substituting $u=2x+1$, as you have found (however, note the new limits are $2\cdot 0 +1 = 1$ and $2\cdot 4 +1=9$). Now use

$$\int x^\alpha\, dx = \frac{x^{\alpha+1}}{\alpha+1}+C,\forall \alpha\neq -1$$

share|improve this answer

Your Answer

 
discard

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.