Use geometry to evaluate the integral from 0 to 6 of the function f of x, dx for the system of equations:
f(x)= 3 for x LESS than or EQUAL to 3 and
equals 6-x for x > 3.
A. 27
B.13.5
C.12
D.10.5
I could really use some help.
Thank you!
To evaluate the integral of the function f(x) over the interval [0, 6], we need to split the integral into two parts based on the given conditions for the function.
First, let's evaluate the integral from 0 to 3:
∫[0 to 3] f(x) dx = ∫[0 to 3] 3 dx
Since f(x) is a constant value of 3 for x ≤ 3, the value of the integral is simply the constant multiplied by the width of the interval:
∫[0 to 3] 3 dx = 3 * (3 - 0) = 9
Next, let's evaluate the integral from 3 to 6:
∫[3 to 6] f(x) dx = ∫[3 to 6] (6 - x) dx
Here, f(x) is given as (6 - x) for x > 3. We can rewrite the integral as follows:
∫[3 to 6] (6 - x) dx = ∫[3 to 6] 6 dx - ∫[3 to 6] x dx
Since 6 is a constant, we can take it outside the integral:
6 * ∫[3 to 6] dx - ∫[3 to 6] x dx
Evaluating the integrals:
6 * (6 - 3) - (1/2) * ((6)^2 - (3)^2)
= 6 * 3 - (1/2) * (36 - 9)
= 18 - (1/2) * 27
= 18 - 13.5
= 4.5
Finally, we add the two results obtained:
9 + 4.5 = 13.5
So, the value of the integral from 0 to 6 of the function f(x) is 13.5.
Therefore, the correct answer is B. 13.5.
To evaluate the integral of the function f(x) from 0 to 6, we need to find the area under the curve of the function within that interval. Since the function has different formulas for different ranges of x, we need to consider those separately.
First, let's consider the range where x is less than or equal to 3. In this range, f(x) is a constant equal to 3. So, we have a rectangle with a base of 3 units (since the width is 3) and a height of 3 units (since the function is constant). The area of this rectangle is (base * height) = (3 * 3) = 9 square units.
Next, let's consider the range where x is greater than 3. In this range, f(x) is given by f(x) = 6 - x. This is a linear function with a negative slope. If we graph this function, we'll see that it's a line that starts at (3,3) and decreases linearly as x increases. The line intersects the x-axis at x = 6.
To find the area under this part of the graph, we need to find the integral of f(x) = 6 - x from x = 3 to x = 6. The integral of this linear function from a to b is the area between the function and the x-axis within that range. The formula for this is ∫(6 - x) dx = [6x - (x^2/2)] evaluated from 3 to 6.
Evaluating the definite integral, we get [6(6) - (6^2/2)] - [6(3) - (3^2/2)] = (36 - 18) - (18 - 4.5) = 18 - 13.5 = 4.5 square units.
Now, to find the total area under the curve, we add the areas from the two separate ranges: 9 + 4.5 = 13.5 square units.
Therefore, the value of the integral from 0 to 6 is 13.5, which corresponds to option B.
a constant 3 so rectangle from 0 to 3
3 times 3 = 9
then slopes down 3 at constant slope from x = 3 to x = 6
so right trinagle of height 3 and base 3
(1/2)(3*3) = 4.5
add