Well Defined Integrals: Proving and Evaluating with Domain Restrictions

[jamesbob]

Here iv to show that integrals are well defined and find their values:

$$a) I = \left\int_{0}{4} \frac{x - 1}{\sqrt{16 - x^2}}dx$$

this one i can show its well defined ok (the domain will be positive as x will be between 0 and 4) but can't find the value. So far my work goes:

$$I = \int_{0}^{4} ( \frac{x}{\sqrt{16 - x^2}} - \frac{1}{\sqrt{16 - x^2}}$$

I know the first part will be arcsin x/4 but i can't do the first part. I began doing substitution but wasn't sure:

$$u = 16 - x^2 \left \frac{du}{dx} = -2x \left So \left xdx = \frac{-du}{2}$$ so the intergral is
$$\frac{-1}{2} \int_{u = 16}^{u = 0} \frac{du}{\sqrt{16 - u^2}}$$
I don't feel this is right. especially the limits i got, theyr impossible right? To get them i said:
$$u = 16 - x^2 \left So \left u = 16 - (0)^2 = 16 \left and \ u = 16 - 4^2 = 0$$

b) $$\int_{0}^{1} (lnx + 2)$$
This one i can do but can't say why its well defined. Is it because the value of lnx will be 0 but the addition of 2 will keep keep it non-zero?

 

[Living_Dog]

Here iv to show that integrals are well defined and find their values:

$$a) I = \left\int_{0}{4} \frac{x - 1}{\sqrt{16 - x^2}}dx$$

this one i can show its well defined ok (the domain will be positive as x will be between 0 and 4) but can't find the value. So far my work goes:

$$I = \int_{0}^{4} ( \frac{x}{\sqrt{16 - x^2}} - \frac{1}{\sqrt{16 - x^2}}$$

I know the first part will be arcsin x/4 but i can't do the first part. I began doing substitution but wasn't sure:

$$u = 16 - x^2 \left \frac{du}{dx} = -2x \left So \left xdx = \frac{-du}{2}$$ so the intergral is
$$\frac{-1}{2} \int_{u = 16}^{u = 0} \frac{du}{\sqrt{16 - u^2}}$$
I don't feel this is right. especially the limits i got, theyr impossible right? To get them i said:
$$u = 16 - x^2 \left So \left u = 16 - (0)^2 = 16 \left and \ u = 16 - 4^2 = 0$$

b) $$\int_{0}^{1} (lnx + 2)$$
This one i can do but can't say why its well defined. Is it because the value of lnx will be 0 but the addition of 2 will keep keep it non-zero?

Try this substitution: $$x = 4sin\theta$$ for both integrals of part (a).

-LD

 

[HallsofIvy]

If u= 16- x², then
$$\sqrt{16-x^2}$$
is NOT
$$\sqrt{16- u^2}$$
!