(13) Quadratic mean ${\displaystyle ={\tfrac {1}{\sqrt {2}}}{\sqrt {A_{1}^{2}+A_{3}^{2}}}}$; arithmetical mean ${\displaystyle =0}$. 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 ${\displaystyle \sin ^{2}x}$ we write

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

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

At the lower limit the substitution of ${\displaystyle 0}$ for ${\displaystyle x}$ causes all this to vanish, whilst at the upper limit the substitution of ${\displaystyle 2\pi }$ for ${\displaystyle x}$ gives ${\displaystyle A_{1}^{2}\pi +A_{3}^{2}\pi }$. And hence the answer follows.

(14) Area is ${\displaystyle 62.6}$ square units. Mean ordinate is ${\displaystyle 10.42}$.

(16) ${\displaystyle 436.3}$. (This solid is pear shaped.)

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

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

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

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

(5) ${\displaystyle \sin(\log _{\epsilon }x)+C}$.

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