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Question Number 131305 by Algoritm last updated on 03/Feb/21

	Answered by EDWIN88 last updated on 03/Feb/21

	$(8) \lim_{x \to 0} \frac{\cos x - \cos (3 x)}{\ln (1 + x^{2})} =$ $\lim_{x \to 0} \frac{(1 - \frac{x^{2}}{2}) - (1 - \frac{9 x^{2}}{2})}{x^{2}} = 4$
	Answered by EDWIN88 last updated on 03/Feb/21

	$(6) f^{'} (1) = \lim_{x \to 0} \frac{f (1 + x) - f (1)}{x}$ $f^{'} (1) = \lim_{x \to 0} \frac{\sqrt{x + 4} - \sqrt{4}}{x}$ $f^{'} (1) = \lim_{x \to 0} [\frac{x}{x [\sqrt{x + 4} + \sqrt{4}]}]$ $= \frac{1}{4}$
	Answered by EDWIN88 last updated on 03/Feb/21
	$(10) ∫_0 ^( 4) ((x^3 dx)/( (√(x^2 +2)))) let x^2 +2 = p^2 ⇒x dx = p dp → determinant (((x=4→p=3(√2))),((x=0→p=(√2)))) E = ∫_(√2) ^( 3(√2)) (((p^2 −2) p dp)/p) E= [(1/3)p^3 −2p ]_(√2) ^(3(√2)) = ((1/3)p [ p^2 −6 ])_(√2) ^(3(√2)) E= (1/3){ 3(√2) . 12−(√2) .(−4) } E = (1/3).40(√2) = ((40(√2))/3)$
	$(10) \int_{0}^{4} \frac{x^{3} d x}{\sqrt{x^{2} + 2}}$ $l e t x^{2} + 2 = p^{2} \Rightarrow x d x = p d p \to \begin{array}{cc} x = 4 \to p = 3 \sqrt{2} \\ x = 0 \to p = \sqrt{2} \end{array}$ $E = \int_{\sqrt{2}}^{3 \sqrt{2}} \frac{(p^{2} - 2) p d p}{p}$ $E = {[\frac{1}{3} p^{3} - 2 p]}_{\sqrt{2}}^{3 \sqrt{2}} = {(\frac{1}{3} p [p^{2} - 6])}_{\sqrt{2}}^{3 \sqrt{2}}$ $E = \frac{1}{3} {3 \sqrt{2} . 12 - \sqrt{2} . (- 4)}$ $E = \frac{1}{3} . 40 \sqrt{2} = \frac{40 \sqrt{2}}{3}$
	Answered by Ar Brandon last updated on 03/Feb/21

	$\underset{n = 1}{\sum^{\infty}} \frac{1}{n (n + 2)} = \frac{1}{2} \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + 2})$ $\frac{1}{2} (1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{6} + \cdot \cdot \cdot) = \frac{1}{2} (1 + \frac{1}{2}) = \frac{3}{4}$
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