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

${\sqrt {{\dfrac {1}{2\pi }}\int _{0}^{2\pi }(A_{1}\sin x+A_{3}\sin 3x)^{2}\,dx}}$.

Now the integration indicated by

$\int (A_{1}^{2}\sin ^{2}x+2A_{1}A_{3}\sin x\sin 3x+A_{3}^{2}\sin ^{2}3x)\,dx$

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

${\dfrac {1-\cos 2x}{2}}$.

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

${\dfrac {1-\cos 6x}{2}}$.

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

${\dfrac {A_{1}^{2}}{2}}\left(x-{\dfrac {\sin 2x}{2}}\right)+A_{1}A_{3}\left({\dfrac {\sin 2x}{2}}-{\dfrac {\sin 4x}{4}}\right)+{\dfrac {A_{3}^{2}}{2}}\left(x-{\dfrac {\sin 6x}{6}}\right)$.

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.)

 

---

 

Exercises XIX. (p. 233.)

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