Properties of integrals Suppose ∫0^3 f(x) d x=2 ∫3^6 f(x) d x=-5, and ∫3^6 g(x) d x=1 . Evaluate

step 1 All right, so we got more integral rules here. So for this first one, I'm actually going to see

Properties of integrals Suppose ∫0^3 f(x) d x=2 ∫3^6 f(x) d...

Properties of integrals Suppose $\int_{0}^{3} f(x) d x=2$ $\int_{3}^{6} f(x) d x=-5,$ and $\int_{3}^{6} g(x) d x=1 .$ Evaluate the following integrals.
a. $\int_{0}^{3} 5 f(x) d x$
b. $\int_{3}^{6}(-3 g(x)) d x$
c. $\int_{3}^{6}(3 f(x)-g(x)) d x$
d. $\int_{6}^{3}(f(x)+2 g(x)) d x$

Key Concepts

Linearity of Integration

This concept states that the integral of a linear combination of functions is equal to the same linear combination of the integrals of those functions. In practical terms, constants can be factored out of the integral and individual integrals can be added or subtracted, which simplifies the evaluation of more complex integrals using known values.

Additivity of Integrals over Adjacent Intervals

Additivity means that if an interval is partitioned into contiguous subintervals, the integral over the entire interval can be obtained by summing the integrals over each subinterval. This property is particularly useful when known integrals over specific subintervals can be combined to provide the value over a larger interval.

Reversal of Limits

This property explains that changing the order of the limits of integration results in a change of sign for the integral. It is an essential aspect of integral calculus, as it enables the evaluation or rearrangement of integrals by recognizing that the integral from a to b is the negative of the integral from b to a.

Transcript

00:01 All right, so we got more integral rules here.

00:02 So for this first one, i'm actually going to see that this five can come out as a constant.

00:06 So we're going to have five times the integral of three, zero, f of x, dx.

00:18 All right, so three to zero f of x, dx is this one.

00:21 So we're going to have five times two for that, and the answer for that will be 10 for a.

00:29 All right, for b, we're going to have the integral from 6 to 3 of negative 3.

00:39 I'm just going to put this negative 3 out here now because it's a constant.

00:42 It could come out of g of x, dx.

00:48 So similar to the last one, we have the 6 to 3 gx, dx over here, and that's equal to 1.

00:54 So we're going to know that this is going to be negative 3 times 1.

00:58 And answer for this one will be negative 3.

01:02 For b.

01:04 For c, i have a little more complex one.

01:07 We're going to have an integral from 6 to 3.

01:10 6 to 3, and it's going to be 3 times f of x minus g of x d x.

01:26 So the best method for this one is going to be to break this integral in 2 because there's a minus sign in between.

01:31 So it's going to look something like this.

01:32 We're going to have a 6 to 3.

01:35 We're actually going to put this 3 out in front because it's a constant.

01:38 So f of x, dx.

01:43 Then we can say minus the integral.

01:46 There's no constant.

01:47 So it's just going to be the same limits...

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