Find the area of the region bounded by the graphs of y = x2 − 4x and

Question

y = x − 4.

a) - 4.500
b) 4.500
c) 2.333
d) None of these

Answer 1

find the graphs' intersections ... x^2 - 4x = x - 4 ... x^2 - 5x + 4 = 0

integrate between the two intersection points

from 1 to 4 , ∫ x^2 - 5x + 4 dx

Answer 2

I knew it until the functions intersect 1,4 but I am confused with the rest.. Please help me with this. thanks in advance

Answer 3

The area consists of a lot of very thin rectangles, of width dx
The height of each rectangle is the distance between the curves. If you draw the graphs, you can see that the parabola lies below the line in the interval [1,4]. So the distance between the curves is
(x - 4) - (x^2 - 4x) = -x^2+5x-4
Now just do the calculus part: take the integral, which sums up all those thin rectangles
A = ∫[1,4] (-x^2+5x-4) dx
= -1/3 x^3 + 5/2 x^2 - 4x [1,4]
= (-1/3 * 4^3 + 5/2 * 4^2 - 4*4)-(-1/3 * 1^3 + 5/2 * 1 - 4*1)
= (-64/3 + 40 - 16) - (-1/3 + 5/2 - 4)
= 8/3 - (-11/6)
= 9/2
When confronted with these area integrals, always draw a picture, and break it up into thin rectangles of width dx (or dy, as needed) and figure each rectangle's height.

Answer 4

evaluate ... (x^3 / 3) - (5/2 x^2) + (4 x) ... at x=4 and x=1

the difference is the area

Answer 5

so wil be 4.5. Answer B, right?

Answer 6

9/2 is about 4.5

Answer 7

Answer: a) -4.5
From 1 to 4 integralx^2 -5x+4dx
Plug it in the calculator and get -4.5.

Answer 8

Sorry, I was wrong.
THE ANSWER IS 4.5
Which is B