For-what-values-of-a-and-b-will-the-integral-a-b-10-x-x-2-dx-be-at-maximum-

Question Number 63372 by necx1 last updated on 03/Jul/19

F o r w h a t v a l u e s o f a a n d b w i l l t h e

i n t e g r a l \int_{a}^{b} \sqrt{10 - x - x^{2}} d x b e a t

m a x i m u m

Commented by mr W last updated on 03/Jul/19

y = \sqrt{10 - x - x^{2}} = \sqrt{{(\frac{\sqrt{41}}{2})}^{2} - {(x + \frac{1}{2})}^{2}}

t h i s i s a s e m i c i r c l e w i t h r a d i u s \frac{\sqrt{41}}{2} a n d

c e n t e r (- \frac{1}{2}, 0)

m a x . v a l u e o f i n t e g r a l i s t h e a r e a

o f s e m i c i r c l e, i . e .

a = - \frac{1}{2} - \frac{\sqrt{41}}{2}

b = - \frac{1}{2} + \frac{\sqrt{41}}{2}

m a x . I = \frac{π}{2} {(\frac{\sqrt{41}}{2})}^{2} = \frac{41 π}{8}

Commented by mathmax by abdo last updated on 03/Jul/19

l e t f (a, b) = \int_{a}^{b} \sqrt{10 - x - x^{2}} d x \Rightarrow f (a, b) = \int_{a}^{b} \sqrt{- (x^{2} + x - 10)} d x

= \int_{a}^{b} \sqrt{- (x^{2} + 2 x \frac{1}{2} + \frac{1}{4} - \frac{1}{4} - 10)} d x = \int_{a}^{b} \sqrt{- {(x + \frac{1}{2})}^{2} + \frac{1}{4} + 10} d x

= \int_{a}^{b} \sqrt{\frac{41}{4} - {(x + \frac{1}{2})}^{2}} d x c h a n g e m e n t x + \frac{1}{2} = \frac{\sqrt{41}}{2} s i n t g i v e

f (a, b) = \int_{\frac{2 a + 1}{\sqrt{41}}}^{\frac{2 b + 1}{\sqrt{41}}} \frac{\sqrt{41}}{2} \sqrt{1 - {s i n}^{2} t} \frac{\sqrt{41}}{2} c o s t d t

= \frac{41}{4} \int_{\frac{2 a + 1}{\sqrt{41}}}^{\frac{2 b + 1}{\sqrt{41}}} {c o s}^{2} t d t = \frac{41}{8} \int_{\frac{2 a + 1}{\sqrt{41}}}^{\frac{2 b + 1}{\sqrt{41}}} (1 + c o s (2 t)) d t

= \frac{41}{8} {\frac{2 b + 1}{\sqrt{41}} - \frac{2 a + 1}{\sqrt{41}}} + \frac{41}{16} {[s i n (2 t)]}_{\frac{2 a + 1}{\sqrt{41}}}^{\frac{2 b + 1}{\sqrt{41}}}

= \frac{\sqrt{41}}{4} (b - a) + \frac{41}{16} {s i n (\frac{4 b + 2}{\sqrt{41}}) - s i n (\frac{4 a + 2}{\sqrt{41}})} l e t s u p p o s e b ⩾ a (b f i x e d)

\frac{\partial f}{\partial a} (a, b) = - \frac{\sqrt{41}}{4} - \frac{41}{16} \frac{4}{\sqrt{41}} c o s (\frac{4 a + 2}{\sqrt{41}}) = - \frac{\sqrt{41}}{4} - \frac{\sqrt{41}}{4} c o s (\frac{4 a + 2}{\sqrt{41}})

= - \frac{\sqrt{41}}{4} {1 + c o s (\frac{4 a + 2}{\sqrt{41}})} = 0 \Rightarrow c o s (\frac{4 a + 2}{\sqrt{41}}) = c o s (π) \Rightarrow

\frac{4 a + 2}{\sqrt{41}} = π + 2 k π o r \frac{4 a + 2}{\sqrt{41}} = - π + 2 k π \Rightarrow

4 a + 2 = \sqrt{41} (2 k + 1) π o r 4 a + 2 = \sqrt{41} (2 k - 1) π \Rightarrow

a = \frac{\sqrt{41} (2 k + 1) π - 2}{4} o r a = \frac{\sqrt{41} (2 k - 1) π - 2}{4} (k \in Z) r e s t t o d r e s z t h e

v a r i a t i o n o f f (a, b) \dots . . b e c o n t i n u e d \dots

Answered by MJS last updated on 03/Jul/19

F (x) = \int \sqrt{10 - x - x^{2}} d x

\underset{a}{\int^{b}} \sqrt{10 - x - x^{2}} d x = F (b) - F (a)

the maximum for a is at

\frac{d}{d a} [F (b) - F (a)] = \sqrt{10 - a - a^{2}} = 0 \Rightarrow a = - \frac{1}{2} \pm \frac{\sqrt{41}}{2}

the maximum for b is similar at

b = - \frac{1}{2} \pm \frac{\sqrt{41}}{2}

b > a \Rightarrow a = - \frac{1}{2} - \frac{\sqrt{41}}{2} \land b = - \frac{1}{2} + \frac{\sqrt{41}}{2}

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