### Video Transcript

Find the area of the region enclosed by the curves π¦ is equal to π₯, π¦ is equal to the sin of π₯, π₯ is equal to π by two, and π₯ is equal to π.

In this question, weβre asked to find the area of the region enclosed by four curves. And whenever weβre asked to find an area of a region enclosed by curves, itβs a good idea to sketch the information weβre given. Weβre going to want to sketch the three lines π¦ is equal to π₯, π₯ is equal to π by two, and π₯ is equal to π and the curve π¦ is equal to the sin of π₯. To do this, itβs usually a good idea to start with the vertical lines, π₯ is equal to π by two and π₯ is equal to π. Doing this, we get the following.

Now, letβs sketch the curve π¦ 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 by two and then an π₯-intercept at π₯ is equal to π. This then gives us a shape which looks somewhat like the following. Finally, we also need to sketch the line π¦ 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 weβre asked to find the area of. Itβs the region bounded above by π¦ is equal to π₯, below by π¦ is equal to the sin of π₯, and to the left by π₯ is equal to π by two and to the right by π₯ is equal to π.

And since π¦ is equal to π₯ and π¦ 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 to π of π of π₯ minus π of π₯ with respect to π₯ is the area under the curve π¦ is equal to π of π₯ and above the curve π¦ is equal to π of π₯ between π₯ is equal to π and π₯ is equal to π. And this is provided π of π₯ and π of π₯ are integrable on this interval and π of π₯ is greater than or equal to π of π₯ for all values of π₯ in the closed interval from π to π. In our case, we want our value of π to be the lower value of π₯, thatβs π by two, and π to be the upper value of π₯, which is π.

And of course we know that π₯ and sin of π₯ are both integrable. In particular, we know that this is true because theyβre both continuous. Finally, we can see in our diagram that π¦ is equal to π₯ is above π¦ 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 π¦-output at the start of this interval. The line π¦ is equal to π₯ will have an output of π by two when π₯ is equal to π by two. However, the sine function has a maximum output of one, which occurs when π₯ is equal to π by two. And of course, π¦ is equal to π₯ is an increasing function. So we can see that over this interval, π¦ is equal to π₯ is above π¦ is equal to sin of π₯.

Therefore, weβre allowed to use this result to find the area of the shaded region. Itβs given by the integral from π by two to π of π₯ minus the sin of π₯ with respect to π₯. We can evaluate this integral term by term. Letβs start with integrating the first term. Thatβs 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 itβs worth pointing out since weβre calculating a definite integral, we donβt 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 πΆ. So when we integrate negative the sin of π₯, weβll get negative one multiplied by this expression. Thatβs the cos of π₯. And again, we need to add a constant of integration πΆ. However, weβre working with a definite integral, so we donβt need this constant.

Therefore, weβve 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 by two and π₯ 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 π₯ 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 thatβs 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, weβve 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 π¦ is equal to π₯, π¦ is equal to the sin of π₯, π₯ is equal to π by two, and π₯ is equal to π is negative one plus three π squared over eight square units.