0-1-1-x-3-1-3-2-dx-

Question Number 53078 by gunawan last updated on 16/Jan/19

\int_{0}^{1} \frac{1}{{(x^{3} + 1)}^{3 / 2}} d x = \dots

Commented by MJS last updated on 17/Jan/19

\approx . 774606

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jan/19

1 + x^{3} > 1 + x^{2}

\frac{1}{1 + x^{3}} < \frac{1}{1 + x^{2}}

b u t i n t h e i n t e r v a l [0, 1]

\frac{1}{{(1 + x^{3})}^{\frac{3}{2}}} > \frac{1}{{(1 + x^{2})}^{\frac{3}{2}} s o}

i a m r e c t i f y i n g

\int_{0}^{1} \frac{d x}{{(1 + x^{3})}^{\frac{3}{2}}} > \int_{0}^{1} \frac{d x}{{(1 + x^{2})}^{\frac{3}{2}}}

n o w

\int_{0}^{1} \frac{d x}{{(1 + x^{2})}^{\frac{3}{2}}}

x = t a n a

\int_{0}^{\frac{π}{4}} \frac{{s e c}^{2} a}{{(1 + {t a n}^{2} a)}^{\frac{3}{2}}} d a

\int_{0}^{\frac{π}{4}} \frac{{s e c}^{2} a d a}{{s e c}^{3} a}

\int_{0}^{\frac{π}{4}} c o s a d a

∣ s i n a ∣_{0}^{\frac{π}{4}}

= \frac{1}{\sqrt{2}}

s o I > \frac{1}{\sqrt{2}}

I > 0.71

a t t a c h i n g g r a p h \dots

a n s w e r y e t t o f i n d \dots .

Commented by tanmay.chaudhury50@gmail.com last updated on 17/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 17/Jan/19

a r e a o f t r a p a z i u m c a l c u l a t i o n

w h e n f (x) = \frac{1}{{(1 + x^{3})}^{\frac{3}{2}}}

a t x = 1 f (1) = \frac{1}{{(2)}^{\frac{3}{2}}} = \frac{1}{2 \sqrt{2}}

a t x = 0 f (0) = \frac{1}{{(1 + 0)}^{\frac{3}{2}}} = 1

s o a r e a o f t r a p a z i u m = \frac{1}{2} [f (0) + f (1)] \times 1

= \frac{1}{2} (1 + \frac{1}{2 \sqrt{2}}) \times 1 = 0.68

f r o m g r a p h i t i s c l e a r t h a t a r e a b o u n d e d i, e

\int_{0}^{1} \frac{d x}{{(1 + x^{3})}^{\frac{3}{2}}} > 0.68