Determine the area of the shaded region in the following figur...

Question

Determine the area of the shaded region in the following figures.

Graph showing the curves of y = sin x and y = cos x, with a shaded area between them from 0 to π/2.

Accepted Answer

Welcome back, everyone. Find the area enclosed by the shaded region in the figure below. A LN of 2, B LN of 3 divided by 2, C, 1 divided by LN of 2, and D LN of 2 divided by 2. What we have to do in this problem is simply recall that the area is simply equal to the integral between two x values of the difference between the upper curve and the lower curve that defines our region. Now let's notice that our region shaded in blue begins at X of 0, right? So we can see that our area is first all equal to the integral from 0 up to some value, right? Let's notice that first of all the upper curve is the blue one up until the intersection point with the red curve, right, and the intersection point occurs at pi divided by 4, meaning we're going to split our region into. Two parts. First of all, we're going to add the integral from 0 to pi divided by 4 of the upper curve minus the lower curve. Now, our lower curve is simply Y equals 0, the x-axis, right? Because our region is bounded by the blue curve Y equals tangent X and the x axis, which has a form of Y equals 0, so we simply take tangent X minus 0, and this means that we're integrating tangent X D X from 0 to pi divided by 4, and this gives us the area of the first region. And then what we want to do is add The integral from pi divided by 4. Up to pi divided by 2. This is where our shaded region ends. Right, and we want to integrate cotangent because now from this point our upper curve is the red one which is cotangent, and the lower curve again is the x axis which is Y equals 0. So we're simply integrating cotangent X minus 0, which is cotangent X. Well, once we have our setup, our area consists of a sum of two integrals. We simply want to integrate each. Now the integral of tangent X is negative Ln of the absolute value of cosine X. We want to evaluate the result from 0 to pi divided by 4. And then we want to add the integral of cotangent XDX. The integral of cotangent is LN. Of the absolute value of sine x and we want to evaluate the result from pi divided by 4 up to pi divided by 2. So now let's go ahead and evaluate each result. For the first part, we get negative LN of the absolute value of cosine. Of 5 divided by 4 we then subtract. LN Of the absolute value of cosine of zero. And now considering the second part, we are adding Ln of the absolute value of sine. Of pi divided by 2. Minus lm of the absolute value of sine of pi divided by 4. Well done, so we have our set up, and now let's valuate each value. We get negative LM. Cosine of pi divided by 4 is square root of 2 divided by 2, we can also drop the absolute value. We then Add LN of cosine of 0. Now, cosine of 0 is 1, so we add LN of 1. Followed by the addition of LN of sine of pi divided by 2, that's Llan of 1 again. And then we subtract Alan of sign of pi divided by 4 witches. A Of square root of 2 divided by 2. Now, what is a of 1? Well, it is 0, so we have two terms that are equal to 0. And we can write our answer as negative tom of square root of 2 divided by 2. Now let's go ahead and simplify this result using the properties of logarithms. We can factor out negative 2. Inparies, we're going to take Alan of square root of 2 minus Alan of 2. And we know that alan of square root of 2 can be written as 1/2 of 2, right, because square root of 2 is to raise to the power of 12, applying the power rule for logarithms, we can bring down 12 and we're subtracting a line of 2. Now let's combine one half of 2 minus Llan of 2 is equal to. Negative 1/2, Helen of 2. And multiplying this value by -2, we get 1. So 1 multiplied by a line of 2 is simply LN of 2, meaning the correct answer to this problem is a. Thank you for watching.

Determine the area of the shaded region in the following figures.

Key Concepts

Definite Integrals and Area Between Curves

Trigonometric Functions and Their Properties

Setting Up the Integral Limits

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