Find-the-area-bounded-by-the-curve-y-3x-2-2x-3-the-x-axis-and-the-line-x-5-and-x-2-

Question Number 32775 by NECx last updated on 02/Apr/18

F i n d t h e a r e a b o u n d e d b y t h e

c u r v e y = 3 x^{2} + 2 x - 3, t h e x a x i s

a n d t h e l i n e x = 5 a n d x = 2

Answered by MJS last updated on 02/Apr/18

1 . check the zeros

x^{2} + \frac{2}{3} x - 1 = 0

x = - \frac{1}{3} \pm \sqrt{\frac{1}{9} + 1} = - \frac{1}{3} \pm \frac{\sqrt{10}}{3}

x_{1} \approx - 1.39 and x_{2} \approx . 72 are both

inside [- 5; 2] \Rightarrow changes of sign

2 . area :

x < x_{1} \Rightarrow f (x) > 0

x \in] x_{1}; x_{2} [\Rightarrow f (x) < 0

x > x_{2} \Rightarrow f (x) > 0

\underset{- 5}{\int^{x_{1}}} f (x) d x - \underset{x_{1}}{\int^{x_{2}}} f (x) d x + \underset{x_{2}}{\int^{2}} f (x) d x =

= F (x) \underset{- 5}{\overset{x_{1}}{∣}} - F (x) \underset{x_{1}}{\overset{x_{2}}{∣}} + F (x) \underset{x_{2}}{\overset{2}{∣}} =

[F (x) = x^{3} + x^{2} - 3 x]

= 91 + \frac{80 \sqrt{10}}{27}

Commented by MJS last updated on 02/Apr/18

no problem, posted a new answer

Commented by NECx last updated on 02/Apr/18

I^{'} m s o s o r r y \dots I m a d e a t y p o e r r o r

t h e l i n e s a r e x = - 5 a n d x = 2 .

Commented by Joel578 last updated on 03/Apr/18

You just set the integral

A = \underset{- 5}{\int^{\frac{- 1 - \sqrt{10}}{3}}} f (x) d x + \underset{\frac{- 1 + \sqrt{10}}{3}}{\int^{\frac{- 1 - \sqrt{10}}{3}}} f (x) d x + \underset{\frac{- 1 + \sqrt{10}}{3}}{\int^{2}} f (x) d x

Commented by MJS last updated on 02/Apr/18

the middle integral is negative,

for the area you need \int ∣ f (x) ∣, so

{it}^{'} s \underset{- 5}{\int^{- \frac{1}{3} - \frac{\sqrt{10}}{3}}} f (x) d x - \underset{- \frac{1}{3} - \frac{\sqrt{10}}{3}}{\int^{- \frac{1}{3} + \frac{\sqrt{10}}{3}}} f (x) d x + \underset{- \frac{1}{3} + \frac{\sqrt{10}}{3}}{\int^{2}} f (x) d x

and {that}^{'} s just what I did

Commented by Joel578 last updated on 03/Apr/18

You are right . {It}^{'} s typo . I have corrected it

Facebook Tweet Pin