Page:Calculus Made Easy.pdf/284

 (13) Quadratic mean $$= \tfrac{1}{\sqrt{2}} \sqrt{A_1^2 + A_3^2}$$; arithmetical mean $$=0$$. The first involves a somewhat difficult integral, and may be stated thus: By definition the quadratic mean will be

Now the integration indicated by

is more readily obtained if for $$\sin^2 x$$ we write

For $$2\sin x \sin 3x$$ we write $$\cos 2x - \cos 4x$$; and, for $$\sin^2 3x$$,

Making these substitutions, and integrating, we get (see p. 202)

At the lower limit the substitution of $$0$$ for $$x$$ causes all this to vanish, whilst at the upper limit the substitution of $$2\pi$$ for $$x$$ gives $$A_1^2 \pi + A_3^2 \pi$$. And hence the answer follows.

(14) Area is $$62.6$$ square units. Mean ordinate is $$10.42$$.

(16) $$436.3$$. (This solid is pear shaped.)

(1) $$\dfrac{x\sqrt{a^2 - x^2}}{2} + \dfrac{a^2}{2} \sin^{-1} \dfrac{x}{a} + C$$.

(2) $$\dfrac{x^2}{2}(\log_\epsilon x - \tfrac{1}{2}) + C$$.

(3) $$\dfrac{x^{a+1}}{a + 1} \left(\log_\epsilon x - \dfrac{1}{a + 1}\right) + C$$.

(4) $$\sin \epsilon^x + C$$.

(5) $$\sin(\log_\epsilon x) + C$$.

(6) $$\epsilon^x (x^2 - 2x + 2) + C$$.