Some fundamental theory of calc 2 questions. I want to evaluate these integrals. I think I understand it, but I just wanted to check:


*

*$$\int_{-1}^1 x^{100} dx$$
$$\frac{1^{101}}{101} - \frac{-1^{101}}{101}$$
$$= \frac{1}{101} + \frac{1}{101} = \frac{2}{101}$$

*$$\int_{1}^9 \sqrt{x} dx$$
$$\frac{2}{3} \cdot 9^{\frac{3}{2}} - \frac{2}{3}$$
$$ = 18 - \frac{2}{3}$$
$$17 \frac{1}{3}$$

*$$\int_{1}^8 x^{\frac{-2}{3}} dx$$
So an antiderivative we can use is: $F(x) = \frac{x^{\frac{1}{3}}}{3}$
$$ \frac{8^{\frac{1}{3}}}{3} - \frac{1^{\frac{1}{3}}}{3}$$
$$ = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$$

*$$\int_{\frac{\pi}{6}}^\pi \sin \theta dx$$
$$ - \cos \pi - -\cos \frac{\pi}{6} $$
$$ = 1 + \frac{\sqrt{3}}{2}$$
$$= \frac{2 + \sqrt{3}}{2}$$

As @littleO points out, the antiderivative of part 3 is incorrect.
$$\int_{1}^8 x^{-2/3} dx = \left[ \frac{x^{1/3}}{1/3} \right]_1^8 = [3\sqrt[3]{x}]_1^8 = 3(2-1) = 3$$
The rest is correct.
