Video Transcript
Find the area of the region enclosed by the curves 𝑦 is equal to 𝑥, 𝑦 is equal to the sin of 𝑥, 𝑥 is equal to 𝜋 by two, and 𝑥 is equal to 𝜋.
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 𝑥, 𝑥 is equal to 𝜋 by two, and 𝑥 is equal to 𝜋 and the curve 𝑦 is equal to the sin of 𝑥. To do this, it’s usually a good idea to start with the vertical lines, 𝑥 is equal to 𝜋 by two and 𝑥 is equal to 𝜋. Doing this, we get the following.
Now, let’s sketch the curve 𝑦 is equal to the sin of 𝑥. We can do this by recalling the sin of 𝑥 starts at the origin has a maximum 𝑥 is equal to 𝜋 by two and then an 𝑥-intercept at 𝑥 is equal to 𝜋. This then gives us a shape which looks somewhat like the following. Finally, we also need to sketch the line 𝑦 is equal to 𝑥. 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 𝑥, below by 𝑦 is equal to the sin of 𝑥, and to the left by 𝑥 is equal to 𝜋 by two and to the right by 𝑥 is equal to 𝜋.
And since 𝑦 is equal to 𝑥 and 𝑦 is equal to the sin of 𝑥 are both integrable, we can find this area by using integration. We can recall the definite integral from 𝑎 to 𝑏 of 𝑓 of 𝑥 minus 𝑔 of 𝑥 with respect to 𝑥 is the area under the curve 𝑦 is equal to 𝑓 of 𝑥 and above the curve 𝑦 is equal to 𝑔 of 𝑥 between 𝑥 is equal to 𝑎 and 𝑥 is equal to 𝑏. And this is provided 𝑓 of 𝑥 and 𝑔 of 𝑥 are integrable on this interval and 𝑓 of 𝑥 is greater than or equal to 𝑔 of 𝑥 for all values of 𝑥 in the closed interval from 𝑎 to 𝑏. In our case, we want our value of 𝑎 to be the lower value of 𝑥, that’s 𝜋 by two, and 𝑏 to be the upper value of 𝑥, which is 𝜋.
And of course we know that 𝑥 and sin of 𝑥 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 𝑥 is above 𝑦 is equal to sin of 𝑥 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 𝑥 will have an output of 𝜋 by two when 𝑥 is equal to 𝜋 by two. However, the sine function has a maximum output of one, which occurs when 𝑥 is equal to 𝜋 by two. And of course, 𝑦 is equal to 𝑥 is an increasing function. So we can see that over this interval, 𝑦 is equal to 𝑥 is above 𝑦 is equal to sin of 𝑥.
Therefore, we’re allowed to use this result to find the area of the shaded region. It’s given by the integral from 𝜋 by two to 𝜋 of 𝑥 minus the sin of 𝑥 with respect to 𝑥. We can evaluate this integral term by term. Let’s start with integrating the first term. That’s 𝑥 or 𝑥 to the first power. We can integrate this by using the power rule for integration. We add one to the exponent of 𝑥 and then divide by this new exponent. We get 𝑥 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 𝑥. We can do this by recalling the integral of the sin of 𝑥 with respect to 𝑥 is negative the cos of 𝑥 plus the constant of integration 𝐶. So when we integrate negative the sin of 𝑥, we’ll get negative one multiplied by this expression. That’s the cos of 𝑥. And again, we need to add a constant of integration 𝐶. 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 𝑥 squared over two plus the cos of 𝑥 evaluated at the limits of integration, 𝑥 is equal to 𝜋 by two and 𝑥 is equal to 𝜋. And now, all we need to do is evaluate our antiderivative at the limits of integration. For the upper limit of integration, we substitute 𝑥 is equal to 𝜋 into the antiderivative. We get 𝜋 squared over two plus the cos of 𝜋. Then, we need to subtract from this the antiderivative evaluated at the lower limit of integration. We need to subtract 𝜋 over two squared over two plus the cos of 𝜋 over two. Now all that’s left to do is to evaluate this expression.
First, the cos of 𝜋 is equal to zero and the cos of 𝜋 by two is equal to one. Next, 𝜋 over two all squared is equal to 𝜋 squared over four. Dividing this by two gives us 𝜋 squared over eight. So we have 𝜋 squared over two minus 𝜋 squared over eight minus one. And we know that 𝜋 squared over two minus 𝜋 squared over eight is equal to three 𝜋 squared over eight. Therefore, we’ve shown the area of the shaded region is negative one plus three 𝜋 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 𝑥, 𝑦 is equal to the sin of 𝑥, 𝑥 is equal to 𝜋 by two, and 𝑥 is equal to 𝜋 is negative one plus three 𝜋 squared over eight square units.
