calculate-f-a-0-1-x-2-ax-1-dx-and-g-a-0-1-xdx-x-2-ax-1-with-a-lt-2-2-find-the-value-of-0-1-x-2-2-x-1-dx-and-0-1-xdx-x-2-2-x-1-

Question Number 74223 by mathmax by abdo last updated on 20/Nov/19

c a l c u l a t e f (a) = \int_{0}^{1} \sqrt{x^{2} + a x + 1} d x a n d g (a) = \int_{0}^{1} \frac{x d x}{\sqrt{x^{2} + a x + 1}}

w i t h ∣ a ∣< 2

2) f i n d t h e v a l u e o f \int_{0}^{1} \sqrt{x^{2} + \sqrt{2} x + 1} d x a n d \int_{0}^{1} \frac{x d x}{\sqrt{x^{2} + \sqrt{2} x + 1}}

Answered by mind is power last updated on 20/Nov/19

f (a) = \int_{0}^{1} \sqrt{{(x + \frac{a}{2})}^{2} + \frac{4 - a^{2}}{4}} d x

l e t (x + \frac{a}{2}) = \sqrt{\frac{4 - a^{2}}{4}} s h (u)

\Rightarrow d x = \sqrt{\frac{4 - a^{2}}{4}} . c h (u) d u

\int_{a r g s h (\frac{a}{\sqrt{4 - a^{2}}})}^{a r g s h (\sqrt{\frac{a + 2}{2 - a}})} . \frac{4 - a^{2}}{4} . {c h}^{2} (u) d u

= \int \frac{4 - a^{2}}{4} . (\frac{c h (2 u) + 1}{2}) d u

= \frac{4 - a^{2}}{4} {[\frac{s h (2 u)}{4} + \frac{u}{2}]}_{a r g s h (\frac{a}{\sqrt{4 - a^{2}}})}^{a r g s h (\sqrt{\frac{a + 2}{2 - a}})}

s h (2 u) = 2 s h (u) c h (u)

s h (2 a r g s h (t)) = 2 t . \sqrt{1 + t^{2}}

= \frac{4 - a^{2}}{4} [\frac{\sqrt{\frac{a + 2}{2 - a}} \sqrt{\frac{4}{2 - a}} - \frac{a}{\sqrt{4 - a^{2}}} \sqrt{\frac{4 - a^{2} + a}{4 - a^{2}}}}{2} + \frac{1}{2} a r g s h (\sqrt{\frac{a + 2}{2 - a}}) - \frac{1}{2} a r g s h (\frac{a}{\sqrt{4 - a^{2}}}) = f (a)

g (a) = 2 f^{'} (a)

2) = f (\sqrt{2}), g (\sqrt{2})

Commented by mathmax by abdo last updated on 20/Nov/19

t h a n k y o u s i r .