Help-please-to-solve-this-If-f-x-1-x-2-for-x-2-2-and-f-x-5-otherwise-Then-what-is-the-value-of-2-2-f-2x-2-dx-

Question Number 116687 by megrex last updated on 05/Oct/20

H e l p p l e a s e, t o s o l v e t h i s \dots

I f f (x) = 1 + x^{2} f o r x \in [- 2, 2] a n d

f (x) = 5 o t h e r w i s e .

T h e n w h a t i s t h e v a l u e o f

\int_{- 2}^{+ 2} f (2 x^{2}) d x ?

Answered by Olaf last updated on 05/Oct/20

f is even and I = \int_{- 2}^{+ 2} f (2 x^{2}) d x = 2 \int_{0}^{2} f (2 x^{2}) d x

2 x^{2} ⩽ 2 \Leftrightarrow x ⩽ 1

In this case f (2 x^{2}) = 1 + 4 x^{4}

f (2 x^{2}) = 5 otherwise .

I = 2 \int_{0}^{1} (1 + 4 x^{4}) d x + 2 \int_{1}^{2} 5 d x

I = 2 (1 + \frac{4}{5}) + 10 = \frac{68}{5}

Commented by megrex last updated on 05/Oct/20

T h a n k y o u O l a f .

Answered by Olaf last updated on 05/Oct/20

other method :

u = 2 x^{2}

d u = 4 x d x = 4 \sqrt{\frac{u}{2}} d x

d x = \frac{d u}{2 \sqrt{2} \sqrt{u}}

I = 2 \int_{0}^{+ 2} f (2 x^{2}) d x = \frac{1}{\sqrt{2}} \int_{0}^{8} f (u) \frac{d u}{\sqrt{u}}

I = \frac{1}{\sqrt{2}} \int_{0}^{2} \frac{1 + u^{2}}{\sqrt{u}} d u + \frac{1}{\sqrt{2}} \int_{2}^{8} \frac{5}{\sqrt{u}} d u

I = \frac{1}{\sqrt{2}} {[2 \sqrt{u} + \frac{2}{5} u^{5 / 2}]}_{0}^{2} + \frac{1}{\sqrt{2}} {[10 \sqrt{u}]}_{2}^{8}

I = 2 + \frac{8}{5} + \frac{1}{\sqrt{2}} (10 \sqrt{8} - 10 \sqrt{2})

I = 2 + \frac{8}{5} + 10 = \frac{68}{5}

Commented by megrex last updated on 05/Oct/20

V e r y n i c e! M a n y t h a n k s .

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