Area between y=x³-2x²+10 and y=3x²+4x-10 from x=-2 to x=2?

Question

I've used the calculator first to find the answer: 160/3. However, when I try to show the work, I can't seem to get that as my answer. It should be as simple as top function minus bottom function. I'm stuck at Int(x³-5x²-4x+20). Appreciate anyone to help show work.

Answer 1

The integration power rule:

∫xⁿ dx = x^(n+1)/(n+1) + C

——————————————————————————————————————

∫[ ( x³ - 2·x² + 10 ) - ( 3x² + 4·x - 10 ) ] dx

= ∫( x³ - 2·x² + 10 - 3x² - 4·x + 10 ) dx

= ∫( x³ - 5·x² - 4·x + 20 ) dx

Use the power rule:

= x^(3+1)/(3+1) - 5·x^(2+1)/(2+1) - 4·x^(1+1)/(1+1) + 20·x^(0+1)/(0+1) + C

= x^(4)/(4) - 5·x^(3)/(3) - 4·x^(2)/(2) + 20·x^(1)/(1) + C

= (1/4)·x^4 - (5/3)·x³ - 2·x² + 20·x + C

——————————————————————————————————————

Thus:

2

∫[ ( x³ - 2·x² + 10 ) - ( 3x² + 4·x - 10 ) ] dx

-2

= [ (1/4)·x^4 - (5/3)·x³ - 2·x² + 20·x ] as x goes from -2 to 2

= [ (1/4)·(2)^4 - (5/3)·(2)³ - 2·(2)² + 20·(2) ] - [ (1/4)·(-2)^4 - (5/3)·(-2)³ - 2·(-2)² + 20·(-2) ]

= [ (2)² - (5/3)·(2)³ - 2³ + 40 ] - [ (2)² - (5/3)·(-2)³ - 2³ - 40 ]

You can cancel factors (the results from even powers will cancel):

= [ -(5/3)·(2)³ + 40 ] - [ (5/3)·(2)³ - 40 ]

= 2·[ -(5/3)·(2)³ + 40 ]

= 2·[ -(40/3) + 40 ]

= 2·[ 80/3 ]

= 160/3

— — — — — — — — — — — — — — — —

or just evaluate:

= [ (2)² - (5/3)·(2)³ - 2³ + 40 ] - [ (2)² - (5/3)·(-2)³ - 2³ - 40 ]

= [ 4 - (40/3) - 8 + 40 ] - [ 4 + (40/3) - 8 - 40 ]

= [ 36 - (40/3) ] - [ -44 + (40/3) ]

= 36 - (40/3) + 44 - (40/3)

= 80 - (80/3)

= 160/3

Answer 2

∫ [(x^3-2x^2+10) - (3x^2+4x-10)] dx , limits [-2,2]

∫ (x^3 -5x^2-4x+20) dx , limits [-2,2]

=x^4/4 -5x^3/3 -4x^2/2 +20x , limits [-2,2]

At x=2, the integral is 22.6666

At x=-2, the integral is -30.6666

Subtracting, 53.3332 (same as 160/3)

160/3 is correct.

Answer 3

Int(x^3-5x^2-4x+20)=(1/4)*x^4 - (5/3)*x^3 - 2x^2 + 20x=F(x)

S=F(+2) - F(-2)=80 - 80/3= 160/3

Answer: 160/3

Answer 4

It's the integral of the difference between the functions:

int(x³ - 5x² - 4x + 20) = ¼x^4 - (5/3)x³ - 2x² + 20x

then evaluate at 2 and -2 and subtract:

[ ¼(16) - (5/3)(8) - 2(4) + 20(2)] - [ ¼(16) - (5/3)(-8) - 2(4) + 20(-2)] =

[ ¼(16) - (5/3)(8) - 2(4) + 20(2)] - [ ¼(16) + (5/3)(8) - 2(4) - 20(2)] =

-80/3 + 80 = 160/3

notice terms with even exponents cancel out (because even powers of 2 and -2 are equal).