∫3,−2 f(x)dx = −4 and ∫4,3 f(x)dx = 3.
Find ∫3,−2 (−4f(x)+3) dx = ______
You cannot plug in f(x) = 4.
All you know is that ∫f(x) dx = 4
∫(-4f(x)+3) dx = ∫-4f(x) dx + ∫3 dx
= -4∫f(x) dx + ∫3 dx
= -4∫f(x) dx + 3x
Now, using the limits [-2,3], that is
= -4(-4) + (3*3 - 3(-2))
= 16 + 15
= 31
I am new to integrals but i assumed that since f(x)dx is -4 and the integral is the same value you can just plug in -4 for f(x) and solve but it was incorrect
To find the value of ∫3,−2 (−4f(x)+3) dx, we can use the linearity property of integrals.
The linearity property states that the integral of a sum or difference of functions is equal to the sum or difference of their integrals. In this case, we have ∫3,−2 (−4f(x)+3) dx, which can be written as ∫3,−2 (−4f(x)) dx + ∫3,−2 (3) dx.
Now, let's solve each integral separately and then add the results together.
For the first integral, ∫3,−2 (−4f(x)) dx, we can use the constant multiple property of integrals, which states that the integral of a constant times a function is equal to the constant times the integral of the function. In this case, the constant is -4.
Using this property, we can rewrite the integral as -4 ∫3,−2 f(x) dx. And we are given that ∫3,−2 f(x) dx = -4. So, substituting this value, we get -4(-4) = 16.
For the second integral, ∫3,−2 (3) dx, we are integrating a constant function, and the integral of a constant times dx is equal to the constant times the measure of the interval. In this case, the constant is 3, and the measure of the interval from 3 to -2 is 3 - (-2) = 5. So, the integral is 3 * 5 = 15.
Finally, we add the results of these two integrals together: 16 + 15 = 31.
Therefore, the value of ∫3,−2 (−4f(x)+3) dx is 31.