let-f-x-0-2-x-t-2-dt-with-x-0-1-calculate-f-x-2-calculate-g-x-0-2-dt-x-t-2-3-find-the-value-of-0-2-4-t-2-dt-and-0-2-dt-3-t-2-4-give-g-x-at-form-of-i

Question Number 66801 by mathmax by abdo last updated on 19/Aug/19

l e t f (x) = \int_{0}^{2} \sqrt{x + t^{2}} d t w i t h x ⩾ 0

1) c a l c u l a t e f (x)

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

3) f i n d t h e v a l u e [o f \int_{0}^{2} \sqrt{4 + t^{2}} d t a n d \int_{0}^{2} \frac{d t}{\sqrt{3 + t^{2}}}

4) g i v e g^{^{'}} (x) a t f o r m o f i n t e g r a l .

Commented by mathmax by abdo last updated on 21/Aug/19

1) f (x) = \int_{0}^{2} \sqrt{x + t^{2}} d t c h a n g e m e n t t = \sqrt{x} u g i v e

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

g i v e f (x) = x \int_{0}^{a r g s h (\frac{2}{\sqrt{x}})} c h (z) c h (z) e z = \frac{x}{2} \int_{0}^{a r g s h (\frac{2}{\sqrt{} x})} (c h (2 z) + 1) d z

= \frac{x}{4} {[s h (2 z)]}_{0}^{l n (\frac{2}{\sqrt{x}} + \sqrt{1 + \frac{4}{x}})} + \frac{x}{2} l n (\frac{2}{\sqrt{x}} + \sqrt{1 + \frac{4}{x}})

= \frac{x}{8} {[e^{2 z} - e^{- 2 z}]}_{0}^{l n (\frac{2 + \sqrt{x + 4}}{\sqrt{x}})} + \frac{x}{2} l n (\frac{2 + \sqrt{x + 4}}{\sqrt{x}}) \Rightarrow

f (x) = \frac{x}{8} {{(\frac{2 + \sqrt{x + 4}}{\sqrt{x}})}^{2} - \frac{1}{{(\frac{2 + \sqrt{x + 4}}{\sqrt{x}})}^{2}}} + \frac{x}{2} l n (\frac{2 + \sqrt{x + 4}}{\sqrt{x}})

2) w e h a v e f^{^{'}} (x) = \int_{0}^{2} \frac{1}{2 \sqrt{x + t^{2}}} d t = \frac{1}{2} g (x) \Rightarrow g (x) = 2 f^{^{'}} (x)

r e s t t o c a l c u l a t e f^{^{'}} (x) . . b e c o n t i n u e d \dots

Commented by mathmax by abdo last updated on 21/Aug/19

3) w e h a v e \int_{0}^{2} \sqrt{4 + t^{2}} d t = f (4) = \frac{1}{2} {{(\frac{2 + 2 \sqrt{2}}{2})}^{2} - \frac{1}{{(\frac{2 + 2 \sqrt{2}}{2})}^{2}}

+ 2 l n (\frac{2 + 2 \sqrt{2}}{2}) = \frac{1}{2} {{(1 + \sqrt{2})}^{2} - \frac{1}{{(1 + \sqrt{2})}^{2}}} + 2 l n (1 + \sqrt{2})

\int_{0}^{2} \frac{d t}{\sqrt{3 + t^{2}}} =_{t = \sqrt{3} u} \int_{0}^{\frac{2}{\sqrt{3}}} \frac{\sqrt{3} d u}{\sqrt{3} \sqrt{1 + u^{2}}} = \int_{0}^{\frac{2}{\sqrt{3}}} \frac{d u}{\sqrt{1 + u^{2}}}

= {[l n (u + \sqrt{1 + u^{2}})]}_{0}^{\frac{2}{\sqrt{3}}} = l n (\frac{2}{\sqrt{3}} + \sqrt{1 + \frac{4}{3}}) = l n (\frac{2 + \sqrt{7}}{\sqrt{3}})

Commented by mathmax by abdo last updated on 21/Aug/19

4) w e h a v e g (x) = \int_{0}^{2} {(x + t^{2})}^{- \frac{1}{2}} d t \Rightarrow g^{^{'}} (x) = \int_{0}^{2} - \frac{1}{2} {(x + t^{2})}^{- \frac{3}{2}} d t

= - \frac{1}{2} \int_{0}^{2} \frac{d t}{(x + t^{2}) \sqrt{x + t^{2}}} .