Integral that converges or diverges?

[tmt1]

I have:

$$\int_{1}^{3} \frac{1}{\sqrt{3 - x}} \,dx$$

I can do $u = \sqrt{3 -x}$, and $du = \frac{1}{2 \sqrt{3 - x}} dx $ and $dx = 2 \sqrt{3 - x} du $. Plug into original equation:

$$\int_{1}^{3} \frac{2 u }{u} \,du$$ and $2 \int_{1}^{3} \,du = 2u = 2 \sqrt{3 - x} + C$

So $(2\sqrt{3 - 3}) - (2\sqrt{3 - 1})$ Or $- 2\sqrt{2}$. Is this correct?

 

[topsquark]

I have:

$$\int_{1}^{3} \frac{1}{\sqrt{3 - x}} \,dx$$

I can do $u = \sqrt{3 -x}$, and $du = \frac{1}{2 \sqrt{3 - x}} dx $ and $dx = 2 \sqrt{3 - x} du $. Plug into original equation:

$$\int_{1}^{3} \frac{2 u }{u} \,du$$ and $2 \int_{1}^{3} \,du = 2u = 2 \sqrt{3 - x} + C$

So $(2\sqrt{3 - 3}) - (2\sqrt{3 - 1})$ Or $- 2\sqrt{2}$. Is this correct?

Actually [math]du = - \frac{1}{2 \sqrt{3 - x}}~dx[/math].

Never never never never write something like [math]dx = 2 \sqrt{3 - x}~du[/math]. Keep the x's and u's on different sides of the equation. Yes you still get dx = 2u du but if you try to substitute the mixed expression into your integrand you can get messed up very quickly. (And, perhaps more to the point, it will confuse the reader.) Finish out the algebra to it's final form.

And you forgot to change the limits of the integration when you did the u-substitution.

Methodwise you did pretty good. (Nod) Just some detail work.

-Dan

 

[MarkFL]

Here's a slightly different approach...We are given:

$$I=\int_1^3 \frac{1}{\sqrt{3-x}}\,dx$$

This is an improper integral, so let's write:

$$I=\lim_{t\to3}\left(\int_1^t \frac{1}{\sqrt{3-x}}\,dx\right)$$

Let:

$$u=3-x\implies du=-dx$$

And we have:

$$I=\lim_{t\to3}\left(\int_{3-t}^2 u^{-\frac{1}{2}}\,du\right)$$

Apply the FTOC:

$$I=2\lim_{t\to3}\left(\left[u^{\frac{1}{2}}\right]_{3-t}^2\right)$$

$$I=2\lim_{t\to3}\left(\sqrt{2}-\sqrt{3-t}\right)=2\sqrt{2}$$