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Question Number 53477 by maxmathsup by imad last updated on 22/Jan/19

l e t f (a) = \int_{0}^{1} \frac{d t}{\sqrt{x + a} + 3}

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

2) f i n d a l s o \int_{0}^{1} \frac{d t}{\sqrt{x + a} {(\sqrt{x + a} + 3)}^{2}}

3) f i n d t h e v a l u e s o f i n t e g r a l s \int_{0}^{1} \frac{d t}{\sqrt{x + 1} + 3} a n d \int_{0}^{1} \frac{d t}{\sqrt{x + 1} {(\sqrt{x + 1} + 3)}^{2}}

Commented by Abdo msup. last updated on 23/Jan/19

1) w e h a v e f (a) = {[2 (\sqrt{x + a} + 3)]}_{0}^{1}

= 2 (\sqrt{a + 1} - \sqrt{a})

2) w e h a v e f^{^{'}} (a) = \int_{0}^{1} \frac{\partial}{\partial a} (\frac{1}{\sqrt{x + a} + 3}) d x

= \int_{0}^{1} - \frac{{(\sqrt{x + a} + 3)}^{,}}{{(\sqrt{x + a} + 3)}^{2}} d x = - \int_{0}^{1} \frac{1}{2 (\sqrt{x + a}) {(\sqrt{x + a} + 3)}^{2}} d x

\Rightarrow \int_{0}^{1} \frac{d x}{(\sqrt{x + a}) {(\sqrt{x + a} + 3)}^{2}} = - 2 f^{^{'}} (a)

= - 4 {(\sqrt{a + 1} - \sqrt{a})}^{^{'}} = - 4 (\frac{1}{2 \sqrt{a + 1}} - \frac{1}{2 \sqrt{a}})

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

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

\int_{0}^{1} \frac{d x}{(\sqrt{x + 1}) {(\sqrt{x + 1} + 3)}^{2}} = - 2 f^{^{'}} (1) = 2 \frac{\sqrt{2} - 1}{\sqrt{2}}

= \sqrt{2} (\sqrt{2} - 1) = 2 - \sqrt{2} .

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19

1) \frac{1}{\sqrt{x + a} + 3} \times ∣ t ∣_{0}^{1} = \frac{1}{\sqrt{x + a} + 3}

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