7.10, #12. \frac{d}{dx} ∫_{x}^{1} ln t dt = −\frac{d}{dx} ∫_{1}^{x} ln t dt = − ln x.

7.10, #19. The definition of erf(x) implies erf′(x) = (2/\sqrt{π})e^{−x^{2}}. Thus

(x erf(x))′ = erf(x) + (2/\sqrt{π})xe^{−x^{2}}.

7.10, #21. Substitute t = \sqrt{2}u, dt = \sqrt{2}du in the integral defining F (x). Then

F (x) = \sqrt{\frac{2}{π}} ∫_{0}^{x} e^{−t^{2}/2} dt = \frac{2}{\sqrt{π}} ∫_{0}^{x/\sqrt{2}} e^{−u^{2}} du = erf(x/\sqrt{2}).

7.Rev, #2. The area of the triangle is

∫_{0}^{b} \frac{hx}{b} dx = \frac{h}{2b}x^{2}|_{0}^{b} = \frac{hb}{2}.

7.Rev, #3. The area of the circle is 4 times the area of the sector in the first quadrant. Thus the area is

4 ∫_{0}^{r} \sqrt{r^{2} − x^{2}} dx = 2x\sqrt{r^{2} − x^{2}} + 2r^{2} arcsin \frac{x}{r}|_{0}^{r} = 2r^{2} \frac{π}{2} = πr^{2}.

The antiderivative is given by formulas 30 and 28 on page 367. 7.Rev, #4. Use the substitution w = 2x, dw = 2 dx. Then

∫ \frac{dx}{\sqrt{1 − 4x^{2}}} = \frac{1}{2} ∫ \frac{dw}{\sqrt{1 − w^{2}}} = \frac{1}{2} arcsin w + C = \frac{1}{2} arcsin 2x + C.

In particular, ∫_{0}^{π/8}(1/\sqrt{1 − 4x^{2}} dx = (0.5) arcsin(π/4) = 0.4517.

7.Rev, #6.

∫_{0}^{2} f(t) dt = ∫_{0}^{1} t^{2} dt + ∫_{1}^{2}(2 − t) dt = \frac{1}{3}t^{3}|_{0}^{1} + (2t − (t^{2}/2))|_{1}^{2}

= \frac{1}{3} + (2 − (3/2)) = \frac{5}{6}.

7.Rev, #11. Use the substitution w = 1 − x, dw = −dx. Then

∫ x\sqrt{1 − x} dx = ∫ −(1 − w)\sqrt{w} dw

= ∫ (w^{3/2} − w^{1/2}) dw = \frac{2}{5}w^{5/2} − \frac{2}{3}w^{3/2} + C

= \frac{2}{5}(1 − x)^{5/2} − \frac{2}{3}(1 − x)^{3/2} + C.

The integral can also be done by parts with u = x v′ = \sqrt{1 − x}. 7.Rev, #16. Use integration by parts with u = (ln x)^{2} and v′ = 1. Then

∫ (ln x)^{2} dx = x(ln x)^{2} − ∫ x(2 ln x)(1/x) dx = x(ln x)^{2} − 2 ∫ ln x dx.

The last integral can also be done by parts or by the tables: ∫ ln x dx = x ln x − x. Thus ∫ (ln x)^{2} dx = x(ln x)^{2} − 2x ln x + 2x + C.

7.Rev, #17. ∫ ln(x^{2}) dx = ∫ 2 ln x dx = 2x ln x − 2x + C.