Question Video: Finding the Area of a Region Bounded by Trigonometric and Linear Functions Mathematics • Higher Education

Find the area of the region enclosed by the curves 𝑦 = π‘₯, 𝑦 = sin π‘₯, π‘₯ = πœ‹/2, and π‘₯ = πœ‹.

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Video Transcript

Find the area of the region enclosed by the curves 𝑦 is equal to π‘₯, 𝑦 is equal to the sin of π‘₯, π‘₯ is equal to πœ‹ by two, and π‘₯ is equal to πœ‹.

In this question, we’re asked to find the area of the region enclosed by four curves. And whenever we’re asked to find an area of a region enclosed by curves, it’s a good idea to sketch the information we’re given. We’re going to want to sketch the three lines 𝑦 is equal to π‘₯, π‘₯ is equal to πœ‹ by two, and π‘₯ is equal to πœ‹ and the curve 𝑦 is equal to the sin of π‘₯. To do this, it’s usually a good idea to start with the vertical lines, π‘₯ is equal to πœ‹ by two and π‘₯ is equal to πœ‹. Doing this, we get the following.

Now, let’s sketch the curve 𝑦 is equal to the sin of π‘₯. We can do this by recalling the sin of π‘₯ starts at the origin has a maximum π‘₯ is equal to πœ‹ by two and then an π‘₯-intercept at π‘₯ is equal to πœ‹. This then gives us a shape which looks somewhat like the following. Finally, we also need to sketch the line 𝑦 is equal to π‘₯. This is a straight line of slope one passing through the origin. And adding this line to our diagram gives us a sketch which looks somewhat like the following. And now, we can mark on this diagram the region we’re asked to find the area of. It’s the region bounded above by 𝑦 is equal to π‘₯, below by 𝑦 is equal to the sin of π‘₯, and to the left by π‘₯ is equal to πœ‹ by two and to the right by π‘₯ is equal to πœ‹.

And since 𝑦 is equal to π‘₯ and 𝑦 is equal to the sin of π‘₯ are both integrable, we can find this area by using integration. We can recall the definite integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ minus 𝑔 of π‘₯ with respect to π‘₯ is the area under the curve 𝑦 is equal to 𝑓 of π‘₯ and above the curve 𝑦 is equal to 𝑔 of π‘₯ between π‘₯ is equal to π‘Ž and π‘₯ is equal to 𝑏. And this is provided 𝑓 of π‘₯ and 𝑔 of π‘₯ are integrable on this interval and 𝑓 of π‘₯ is greater than or equal to 𝑔 of π‘₯ for all values of π‘₯ in the closed interval from π‘Ž to 𝑏. In our case, we want our value of π‘Ž to be the lower value of π‘₯, that’s πœ‹ by two, and 𝑏 to be the upper value of π‘₯, which is πœ‹.

And of course we know that π‘₯ and sin of π‘₯ are both integrable. In particular, we know that this is true because they’re both continuous. Finally, we can see in our diagram that 𝑦 is equal to π‘₯ is above 𝑦 is equal to sin of π‘₯ over this entire interval. However, sketches can be slightly inaccurate. So it could be worth checking that this is true. To do this, we can find the 𝑦-output at the start of this interval. The line 𝑦 is equal to π‘₯ will have an output of πœ‹ by two when π‘₯ is equal to πœ‹ by two. However, the sine function has a maximum output of one, which occurs when π‘₯ is equal to πœ‹ by two. And of course, 𝑦 is equal to π‘₯ is an increasing function. So we can see that over this interval, 𝑦 is equal to π‘₯ is above 𝑦 is equal to sin of π‘₯.

Therefore, we’re allowed to use this result to find the area of the shaded region. It’s given by the integral from πœ‹ by two to πœ‹ of π‘₯ minus the sin of π‘₯ with respect to π‘₯. We can evaluate this integral term by term. Let’s start with integrating the first term. That’s π‘₯ or π‘₯ to the first power. We can integrate this by using the power rule for integration. We add one to the exponent of π‘₯ and then divide by this new exponent. We get π‘₯ squared divided by two. And it’s worth pointing out since we’re calculating a definite integral, we don’t need a constant of integration. We just need any antiderivative.

Next, we need to integrate negative the sin of π‘₯. We can do this by recalling the integral of the sin of π‘₯ with respect to π‘₯ is negative the cos of π‘₯ plus the constant of integration 𝐢. So when we integrate negative the sin of π‘₯, we’ll get negative one multiplied by this expression. That’s the cos of π‘₯. And again, we need to add a constant of integration 𝐢. However, we’re working with a definite integral, so we don’t need this constant.

Therefore, we’ve shown that the area of the shaded region is π‘₯ squared over two plus the cos of π‘₯ evaluated at the limits of integration, π‘₯ is equal to πœ‹ by two and π‘₯ is equal to πœ‹. And now, all we need to do is evaluate our antiderivative at the limits of integration. For the upper limit of integration, we substitute π‘₯ is equal to πœ‹ into the antiderivative. We get πœ‹ squared over two plus the cos of πœ‹. Then, we need to subtract from this the antiderivative evaluated at the lower limit of integration. We need to subtract πœ‹ over two squared over two plus the cos of πœ‹ over two. Now all that’s left to do is to evaluate this expression.

First, the cos of πœ‹ is equal to zero and the cos of πœ‹ by two is equal to one. Next, πœ‹ over two all squared is equal to πœ‹ squared over four. Dividing this by two gives us πœ‹ squared over eight. So we have πœ‹ squared over two minus πœ‹ squared over eight minus one. And we know that πœ‹ squared over two minus πœ‹ squared over eight is equal to three πœ‹ squared over eight. Therefore, we’ve shown the area of the shaded region is negative one plus three πœ‹ squared over eight square units, which is our final answer.

Therefore, we were able to show the area of the region enclosed by the curves 𝑦 is equal to π‘₯, 𝑦 is equal to the sin of π‘₯, π‘₯ is equal to πœ‹ by two, and π‘₯ is equal to πœ‹ is negative one plus three πœ‹ squared over eight square units.

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