# Improper Integral

## Homework Statement

$$\int_{-3}^{1}\frac{x}{\sqrt{9-x^{2}}}$$

## Homework Equations

Let f be continuous on the half-open interval (a, b] and suppose that
$$\lim_{x \to a^{+}} |f(x)| = \infty$$. Then

$$\int_{a}^{b}f(x) dx = \lim_{ t \to a^{+}}\int_{t}^{b}f(x) dx$$

## The Attempt at a Solution

$$\int_{-3}^{1}\frac{x}{\sqrt{9-x^{2}}} = \lim_{ t \to -3^{+}}\int_{t}^{1}\frac{x}{\sqrt{9-x^{2}}}$$

$$u = 9 - x^{2}$$
$$du = -2x dx$$

$$\lim_{ t \to -3^{+}}\int_{t}^{1}\frac{x}{\sqrt{9-x^{2}}} = \lim_{ t \to 0^{+}}-\frac{1}{2}\int_{t}^{8} u^{-1/2} du =\lim_{ t \to 0^{+}}-u^{1/2}|_{t}^{8} =-\sqrt{3sin(1)} + \lim_{ t \to -3^{+}}\sqrt{3sin(t)} =-1.588840129 + ?$$

I get 0 for the limit, but according to Maple and my graphing calculator, that does not give the correct value for this integral. The correct value is approximately -2.8. May I please have some guidance as to what may have went wrong?

$$\int_{-3}^{1}\frac{x}{\sqrt{9-x^{2}}} = \lim_{ t \to -3^{+}}\int_{t}^{1}\frac{x}{\sqrt{9-x^{2}}}$$

$$u = 9 - x^{2}$$
$$du = -2x dx$$

$$\lim_{ t \to -3^{+}}\int_{t}^{1}\frac{x}{\sqrt{9-x^{2}}} = \lim_{ t \to 0^{+}}-\frac{1}{2}\int_{t}^{8} u^{-1/2} du =\lim_{ t \to 0^{+}}-u^{1/2}|_{t}^{8} =-\sqrt{3sin(1)} + \lim_{ t \to -3^{+}}\sqrt{3sin(t)} =-1.588840129 + ?$$

I get 0 for the limit, but according to Maple and my graphing calculator, that does not give the correct value for this integral. The correct value is approximately -2.8. May I please have some guidance as to what may have went wrong?

I don't understand that step where you introduce sine. You should just get:

$$\lim_{ t \to 0^{+}}-u^{1/2}|_{t}^{8} =-\lim_{t \to 0^+}(\sqrt{8} - \sqrt{t}) = 0 - 2\sqrt{2} = -2\sqrt{2} \approx -2.8$$

I don't understand that step where you introduce sine. You should just get:

Oh, god, I made an incredibly stupid mistake....Thank you.

haha, it happens. no worries.