Using properties of integrals Use the value of the first...

Using properties of integrals Use the value of the first integral I to evaluate the two given integrals. $$\int_{0}^{\pi / 2}(2 \sin \theta-\cos \theta) d \theta$$ $$\text { a. } \int_{0}^{\pi / 2}(2 \sin \theta-\cos \theta) d \theta \quad \text { b. } \int_{\pi / 2}^{0}(4 \cos \theta-8 \sin \theta) d \theta$$

Using properties of integrals Use the value of the first integral I to evaluate the two given integrals.
$$\int_{0}^{\pi / 2}(2 \sin \theta-\cos \theta) d \theta$$
$$\text { a. } \int_{0}^{\pi / 2}(2 \sin \theta-\cos \theta) d \theta \quad \text { b. } \int_{\pi / 2}^{0}(4 \cos \theta-8 \sin \theta) d \theta$$

Key Concepts

Fundamental Theorem of Calculus

This theorem links the concept of the antiderivative with the evaluation of definite integrals. It allows the integral of a function over an interval to be computed by finding an antiderivative and subtracting its values at the endpoints, which is essential in determining the exact value of the given integrals.

Reversal of Integration Limits

Changing the order of the limits in a definite integral results in a sign change of the integral. This property is useful for relating integrals with reversed limits back to standard form, thereby simplifying their evaluation.

Linearity of Integration

This property states that the integral of a sum or difference of functions is equal to the sum or difference of the integrals of those functions, and that constant factors can be moved outside of the integral. It allows complex integrals to be broken into simpler parts which are easier to evaluate.

Transcript

00:04 We want to calculate this definite integral and then use the result to calculate this definite integral.

00:12 The integral of 2 -sign theta, derivative of sine theta, i'm sorry, the integral of sine theta will be negative cosine theta.

00:21 So negative 2 cosine theta minus the integral of cosine, which is sine theta.

00:32 We have to evaluate this between pi over 2 and 0.

00:40 So, all right, double -checking all work.

00:42 Derivative of cosine is negative sign, so negative two times negative sign is two -sign.

00:47 Derivative of sine is cosine, so that would be the negative of cosine.

00:51 That's what we'll be had.

00:52 All right.

00:54 We'll come right back and evaluate this.

01:00 Okay.

01:03 Evaluating this expression, when data is pi over two, the cosine of pi over two is zero, minus the sine of pi over 2, which is 1.

01:19 And then we have to subtract this same expression you evaluated now when theta is 0.

01:27 Cosign of 0 is 1.

01:28 So negative 2 times 1 is negative 2 minus the sign of 0.

01:38 So we have negative 1 subtract.

01:53 Negative 2 minus 0 is negative 2.

01:57 So negative 1 subtract negative 2.

01:59 Same thing as negative 1 plus 2 is 1.

02:07 This integral is very similar to this integral.

02:15 First of all, notice the limits of integration are switched.

02:19 It says 0 to pi over 2 is pi over 2 to 0.

02:23 So the very first thing we can do is we know that this will equal switching these limits of integration, 0 on the bottom, pi over 2 up top.

02:41 The integral of this expression from 0 to pi over 2, i'm sorry, let me re -say that.

02:49 The integral of this expression from pi over 2 to 0 will be the negative of the integral of this expression from 0 to pi over 2.

03:00 In other words, if we keep the function the same, four cosine theta minus eight sine theta.

03:15 So the function is the same, but we're integrating from zero to pi over two.

03:24 Okay.

03:25 This function integrated from pi over two to zero will equal the negative of the integral of that function from zero to pie over two.

03:35 Okay, all we did was switch the limits of integration.

03:38 Okay, if we switch the limits of integration, then you're going to get the negative of what you would have...

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