Math Question and Answers


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 126073 by benjo_mathlover last updated on 17/Dec/20

Commented by benjo_mathlover last updated on 17/Dec/20

$T h e g r a p h o f t h e d i f f e r e n t i a b l e$ $f u n c t i o n g w i t h d o m a i n - 6 ⩽ x ⩽ 2 i s$ $s h o w n i n t h e f i g u r e a b o v e . T h e a r e a s$ $o f t h e r e g i o n s b o u n d e d b y t h e x - a x i s$ $a n d t h e g r a p h o f g o n t h e i n t e r v a l s [- 6, - 5]$ $[- 5, - 3], [- 3, 1] a n d [1, 2] a r e 9, 17$ $42 a n d 6 r e s p e c t i v e l y . T h e g r a p h o f g$ $h a s h o r i z o n t a l t a n g e n t a t x = - 4,$ $x = - 3 a n d x = - 1 . L e t h b e t h e$ $f u n c t i o n d e f i n e d b y h (x) = \int_{- 3}^{x} g (t) d t$ $f o r - 6 ⩽ x ⩽ 2 . F i n d t h e v a l u e o f$ $(1) h (1)$ $(2) h (- 6)$ $(3) t h e x - c o o r d i n a t e o f e a c h c r i t i c a l$ $p o i n t o f h o n t h e i n t e r v a l - 6 ⩽ x ⩽ 2$
Commented by liberty last updated on 17/Dec/20

$(1) h (1) = \int_{- 3}^{1} g (t) d t; s i n c e g (t) ⩽ 0 o n t h e$ $i n t e r v a l - 3 ⩽ t ⩽ 1; s o h (1) = - [a r e a o f t h e r e g i o n$ $b o u n d e d b y t h e x - a x i s o n [- 3, 1]$ $h (1) = - 42 .$ $(2) h (- 6) = \int_{- 3}^{- 6} g (t) d t = - \int_{- 6}^{- 3} g (t) d t$ $h (- 6) = - [\int_{- 6}^{- 5} g (t) d t + \int_{- 5}^{- 3} g (t) d t]$ $= - [9 + (- 17)] = 8$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum