Let f(x) be the function defined by f(x) = {x, x≤0
x+1, x>0
The value of the integral of x(f(x))dx from -2 to 1 =?
Well, the integral of x(f(x))dx from -2 to 1 can be calculated by splitting the integral into two parts: one for x ≤ 0 and one for x > 0.
For x ≤ 0, the function f(x) equals x. So the integral of x(f(x))dx from -2 to 0 would be the integral of x*x dx from -2 to 0.
To evaluate this integral, we can use the power rule of integration. The integral of x^n dx, where n is any real number except -1, is (1/(n+1)) * x^(n+1).
Using the power rule, the integral of x*x dx would be (1/3) * x^3.
Evaluating this from -2 to 0, we get [(1/3) * 0^3] - [(1/3) * (-2)^3], which simplifies to 0 - (-8/3) = 8/3.
Now, for x > 0, the function f(x) equals x + 1. So the integral of x(f(x))dx from 0 to 1 would be the integral of x*(x+1) dx from 0 to 1.
Expanding, we have x^2 + x. Again, using the power rule, the integral of x^2 + x dx would be (1/3) * x^3 + (1/2) * x^2.
Evaluating this from 0 to 1, we get [(1/3) * 1^3 + (1/2) * 1^2] - [(1/3) * 0^3 + (1/2) * 0^2], which simplifies to 1/3 + 1/2 = 5/6.
Now, adding the two results together, we get 8/3 + 5/6 = 16/6 + 5/6 = 21/6, which simplifies to 7/2.
Therefore, the value of the integral of x(f(x))dx from -2 to 1 is 7/2.
To find the value of the integral of x(f(x))dx from -2 to 1, we need to split the integral into two parts based on the definition of the function f(x).
For x ≤ 0, the function f(x) is equal to x. So the integral of x(f(x))dx in this range is:
∫[x ≤ 0] x(x) dx
Simplifying this, we get:
∫[x ≤ 0] x^2 dx
Integrating x^2 with respect to x, we get:
[x^3/3] from -2 to 0
Plugging the limits into the antiderivative, we have:
[(0)^3/3] - [(-2)^3/3] = 0 - (-8/3) = 8/3
For x > 0, the function f(x) is equal to x + 1. So the integral of x(f(x))dx in this range is:
∫[x > 0] x(x + 1) dx
Expanding the integrand, we have:
∫[x > 0] (x^2 + x) dx
Integral of x^2 gives:
[x^3/3] from 0 to 1
Integral of x gives:
[x^2/2] from 0 to 1
Plugging the limits into the antiderivative, we have:
[(1)^3/3 + (1)^2/2] - [(0)^3/3 + (0)^2/2] = 1/3 + 1/2 = 5/6
To find the total value of the integral from -2 to 1, we need to sum up the results from the two ranges:
Total value = ∫[x ≤ 0] x^2 dx + ∫[x > 0] (x^2 + x) dx
Total value = 8/3 + 5/6
Total value = (16 + 5)/6
Total value = 21/6
Simplifying, we get:
Total value = 7/2
Therefore, the value of the integral of x(f(x))dx from -2 to 1 is 7/2.
To find the value of the integral of x(f(x))dx from -2 to 1, we need to evaluate the integral using the given function f(x) and its definition for different intervals.
First, let's split the interval of integration [-2, 1] into two parts based on where the function f(x) changes its definition.
For values of x ≤ 0, the function f(x) is equal to x. So, the integral over this interval will be:
∫[x(f(x))]dx from -2 to 0 = ∫[x * x]dx from -2 to 0
Now, let's evaluate the integral of x^2 from -2 to 0:
∫[x^2]dx from -2 to 0 = [-1/3 * x^3] from -2 to 0
Plugging in the values, we get:
[-1/3 * 0^3] - [-1/3 * (-2)^3] = -1/3 * 0 - (-8/3) = -8/3
Next, for values of x > 0, the function f(x) is equal to x + 1. So, the integral over this interval will be:
∫[x(f(x))]dx from 0 to 1 = ∫[(x + 1) * x]dx from 0 to 1
Now, let's evaluate the integral of (x + 1) * x from 0 to 1:
∫[(x + 1) * x]dx from 0 to 1 = [(1/3) * x^3 + (1/2) * x^2] from 0 to 1
Plugging in the values, we get:
[(1/3) * 1^3 + (1/2) * 1^2] - [(1/3) * 0^3 + (1/2) * 0^2] = 1/3 + 1/2 = 5/6
Finally, to find the value of the integral of x(f(x))dx from -2 to 1, we sum up the integrals over the different intervals:
∫[x(f(x))]dx from -2 to 1 = ∫[x * x]dx from -2 to 0 + ∫[(x + 1) * x]dx from 0 to 1
= -8/3 + 5/6
To simplify, we can find a common denominator:
= (-16/6) + (5/6)
Combining the terms, we get:
= -16/6 + 5/6
= -11/6
Therefore, the value of the integral of x(f(x))dx from -2 to 1 is -11/6.