Find the area between the curve...?

Question

Find the answer to 2 decimal places

y=x^2-5 x
and the x-axis for x between 1 and 8.

(My method was to find the antiderivative and then plug in 8 and 1 into the equation and subtract, resulting in the answer 12.83, but this was wrong)

Answer 1

Ok, let's solve using integral calculus :

Area = integrate(x^2 - 5x)dx

x goes from : x = 1 to x = 8

Area = x^3 / 3 - 5*x^2 / 2

Area = (8^3 / 3 - 320 / 2) - ( 1/3 - 5/2)

Area = 10.6 - ( -2.16 ) = 12.76

Hope that helps

Answer 2

There are 2 regions of the curve, one below x-axis, one above x-axis. The area below the x-axis is a negative value if u just integrate like that. And that should be the cause of why u are not getting the correct answer.

At x-axis, y=0,
x^2 - 5x = 0
x(x-5)=0
x=0 and x=5

Becos the coefficient of the x^2 is a +1, hence this is a minimum graph (U-shaped curve). The region of 05 is above the x-axis.

Hence,
Area between curve & graph for 1 = ∫ -(x^2 - 5x) dx ------- take note of the "-ve", this is added to the normal integral to find area UNDER the x-axis.
= ∫ -x^2 + 5x dx
= [ -x^3/3 + (5/2)*x^2 ]
= [ -(5^3)/3 + (5/2)*(5^2) ] - [ -1/3 +5/2 ]
= [ -125/3 + (5/2)*25 ] - [ -2/6 + 15/6]
= [ -125/3 + 125/2] - (13/6)
= [ -250/6 + 375/6] - 13/6
= 125/6 - 13/6
= 112/6
= 56/3

Area between curve & graph for 5 = ∫ (x^2 - 5x) dx
= ∫ x^2 - 5x dx
= [ x^3/3 - (5/2)*x^2 ]
= [ (8^3)/3 - (5/2)*(8^2) ] - [ (5^3)/3 - (5/2)*(5^2) ]
= [ 512/3 - (5/2)*64 ] - [ 125/3 - (5/2)*25]
= [ 512/3 - 160] - [ 125/3 - 125/2]
= (32/3) - [ 250/6 - 375/6]
= 32/3 - (-125/6)
= 32/3 + 125/6
= 31.5

Total area
= 31.5 + 56/3
= 501/6
= 50.17 (Corrected to 2 dec. places)

Answer 3

x^2 - 5x = 0 when x = 0 and x = 5.

You have to split the region between 1 and 8 into 2 areas. The area between 1 and 5 is beneath the x-axis, so when you integrate the function from 1 to 5 take the opposite of your answer to that part. Then add the integral from 5 to 8.

You get:
-[(5^3/3 - 5*5^2/2)- (1^3/3 - 5*1^2/2)] + (8^3/3 - 5*8^2/2) - (5^3/3 - 5*5^2/2)

= -[125/3 - 125/2) - (1/3 - 5/2)] + (512/3 - 320/2) - (125/3 - 125/2)

= -(124/3 - 120/2) + (387/3 - 195/2)
= 50.17

Answer 4

The way to find the area between the curve from x = 1 to x = 8 is to integrate the function from x = 1 to 8.

The integral of y = x^2-5x is

f(x) = (1/3)x^3 - (5/2)x^2
f(8) - f(1) = (1/3)*8^3 - (5/2)*8^2 - (1/3)*1^2 + (5/2)*1^2
f(8) - f(1) = 12.83

Are you sure that's the wrong answer?

Answer 5

HERES THE ANSWER- TRUST ME!!! What you have to do is set up an integral. Since this curve is part way above the line and partway below the axis, use 2 integrals. From 1 to 5, use -1 * integral(x^2-5x), so you get a positive number. Add this to the integral from 5 to 8 of the same curve. I get 81.67.

Answer 6

y = 4x y = 2x² First, set the two curves equivalent to discover their x-intercepts: 4x = 2x² via commentary, x = 0. as nicely: x = 4/2 = 2 for this reason, the bounds of integration are x = 0 to 2. A = ? 4x - 2x² dx A = [2x² - (2/3)x³] evaluated from x = 0 to 2 A = [2(2)² - (2/3)(2)³] - [2(0)² - (2/3)(0)³] = 8 - (2/3)(8) = 8/3 gadgets²