Find the area of the region bounded by the graphs of y = x2 − 4x and
y = x − 4.
a) - 4.500
b) 4.500
c) 2.333
d) None of these
Sorry, I was wrong.
THE ANSWER IS 4.5
Which is B
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.
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
I knew it until the functions intersect 1,4 but I am confused with the rest.. Please help me with this. thanks in advance
evaluate ... (x^3 / 3) - (5/2 x^2) + (4 x) ... at x=4 and x=1
the difference is the area
so wil be 4.5. Answer B, right?
9/2 is about 4.5
Answer: a) -4.5
From 1 to 4 integralx^2 -5x+4dx
Plug it in the calculator and get -4.5.
To find the area of the region bounded by the two graphs, we need to determine the points where they intersect.
First, let's set the two equations equal to each other:
x^2 - 4x = x - 4
Simplifying, we get:
x^2 - 5x + 4 = 0
Now we can solve this quadratic equation for x. Factoring it, we have:
(x - 4)(x - 1) = 0
Setting each factor equal to zero gives us two solutions:
x - 4 = 0, which gives x = 4
x - 1 = 0, which gives x = 1
So the two graphs intersect at x = 4 and x = 1.
To find the area bounded by the graphs, we need to integrate the absolute difference between the two functions in the given interval.
Let's integrate the absolute difference between the two functions from x = 1 to x = 4:
∫[1,4] |(x^2 - 4x) - (x - 4)| dx
Simplifying, we have:
∫[1,4] |x^2 - 5x + 4| dx
To integrate the absolute value function, we can break the interval into two parts where the function is positive and negative.
For x in [1,4], the expression inside the absolute value is positive, so the integral becomes:
∫[1,4] (x^2 - 5x + 4) dx
Performing the integration, we get:
∫[1,4] (x^2 - 5x + 4) dx = [(1/3)x^3 - (5/2)x^2 + 4x] [1,4]
Plugging in the bounds, we have:
[(1/3)*(4)^3 - (5/2)*(4)^2 + 4*(4)] - [(1/3)*(1)^3 - (5/2)*(1)^2 + 4*(1)]
Simplifying further gives us:
(64/3 - 40 + 16) - (1/3 - 5/2 + 4)
Finally, calculating the result gives us:
(68/3) - (7/6) = (136/6) - (7/6) = 129/6 = 21.5
So the area of the region bounded by the graphs is 21.5.
None of the given options (a, b, c, d) matches the correct answer.