m≤∫_1 ^3 ((1 )/(√(2x^2 +7 ))) .dx≤k find the value of the constant m and k


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Question Number 212635 by Nadirhashim last updated on 19/Oct/24

$m ⩽ \underset{1}{\int^{3}} \frac{1}{\sqrt{2 x^{2} + 7}} . d x ⩽ k f i n d$ $t h e v a l u e o f t h e c o n s t a n t$ $m a n d k$
Commented by Ghisom last updated on 19/Oct/24

$this is a strange question$ $\underset{1}{\int^{3}} \frac{d x}{\sqrt{2 x^{2} + 7}} = \sqrt{2} \ln (1 + 2 \sqrt{2}) - \frac{\sqrt{2}}{2} \ln 7$ $\Rightarrow m = k = \sqrt{2} \ln (1 + 2 \sqrt{2}) - \frac{\sqrt{2}}{2} \ln 7$ $if m, k \in Z \Rightarrow m = 0, k = 1$
	Answered by mehdee7396 last updated on 19/Oct/24

	$f (x) = \frac{1}{\sqrt{2 x^{2} + 7}} \Rightarrow f^{'} (x) = \frac{- 2 x}{(2 x^{2} + 7) \sqrt{2 x^{2} + 7}} < 0 \Rightarrow f ↘$ $f (1) = \frac{1}{3} = m a x & f (3) = \frac{1}{5} = m i n$ $\Rightarrow \frac{2}{5} < \int_{1}^{3} \frac{1}{\sqrt{2 x^{2} + 7}} d x < \frac{2}{3}$ $\Rightarrow m = \frac{2}{5} & k = \frac{2}{3}$
	Commented by Ghisom last updated on 19/Oct/24

	$this is also a strange answer$ $why not$ $m = \frac{13}{25} \land k = \frac{53}{100}$ $or$ $m = \frac{163297}{312500} \land k = \frac{1045101}{2000000}$
	Commented by mehdee7396 last updated on 20/Oct/24

	Commented by Ghisom last updated on 20/Oct/24

	$but this is a special definition, never$ $heard of it before . it makes sense for an$ $integral we cannot solve exactly .$
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