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 ∫


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

$H e l p p l e a s e, t o s o l v e t h i s . . .$ $I f f (x) = 1 + x^{2} f o r x \in [- 2, 2] a n d$ $f (x) = 5 o t h e r w i s e .$ $T h e n w h a t i s t h e v a l u e o f$ $\int_{- 2}^{+ 2} f (2 x^{2}) d x ?$
	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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