### Video Transcript

Find the area of the region above the π₯-axis bounded by the curve π¦ is equal to three times π to the power of π₯ over seven and the lines π₯ is equal to zero and π₯ is equal to 14. Give an exact answer.

We need to find the area of the region bounded by the following lines. Our region lies above the π₯-axis, below the curve π¦ is equal to three π to the power of π₯ over seven, and between the lines π₯ is equal to zero and π₯ is equal to 14. We need to give our answer exactly.

Letβs start by sketching our region. First, weβre told our region lies above the π₯-axis and between the lines π₯ is equal to zero and π₯ is equal to 14. So, weβll mark this on our diagram. Next, we want to sketch our curve π¦ is equal to three π to the power of π₯ over seven.

To do this, we can see itβs a series of transformations from our exponential function. We can see we multiply the entire function by three. This means itβs a vertical stretch with a factor of three. Next, we can see we multiply π₯ by one-seventh. This means itβs a horizontal stretch of factor seven. The important thing here is to note that it will have a similar shape to the function π¦ is equal to π to the power of π₯. So, we can now mark our region π
on the diagram.

And now, we can see we need to find the area under a curve which is fully above the π₯-axis and our curve is defined by a continuous function. This means we can do this by using integration. We recall, for a continuous function π of π₯, if π of π₯ is greater than or equal to zero for values of π₯ greater than or equal to π and less than or equal to π, then the integral from π to π of π of π₯ with respect to π₯ is the area below the curve π¦ is equal to π of π₯, above the π₯-axis, between the lines π₯ is equal to π and π₯ is equal to π.

And we can see this is exactly the situation weβre in in this question. In our case, we want to find the area under the curve π¦ is equal to three π to the power of π₯ over seven between the lines π₯ is equal to zero and π₯ is equal to 14. So, we can do this by calculating the definite integral from zero to 14 of three π to the power of π₯ over seven with respect to π₯. To evaluate this integral, we need to recall our rules for integrating exponential functions.

We recall for any real constants π and π, where π is not equal to zero, the integral of π times π to the power of ππ₯ with respect to π₯ is equal to π times π to the power of ππ₯ divided by π plus a constant of integration πΆ. Of course, in this case, weβre calculating a definite integral. So, we donβt need our constant of integration. In our case, our value of π is three, and our value of π is one over seven. So, we get three π to the power of π₯ over seven all divided by one over seven evaluated at the limits of integration π₯ is equal to zero and π₯ is equal to 14.

Of course, instead of dividing by the fraction one over seven, we can just multiply by seven. Of course, three times seven is equal to 21. So, we get 21π to the power of π₯ over seven evaluated at the limits of integration zero and 14. Now, all we need to do is evaluate this at the limits of integration. Evaluating this at the limits of integration, we get 21π to the power of 14 over seven minus 21π to the power of zero over seven. And we can now simplify this expression.

First, π to the power of zero over seven is just π to the power of zero. But any number raised to the power of zero is just equal to one. Next, we have 14 over seven simplifies to give us two. So, this expression simplifies to give us 21π squared minus 21. Weβll simplify this one more way. Weβll take out the common factor of 21. And this gives us 21 times π squared minus one. Therefore, by using integration, we were able to show the area of the region above the π₯-axis bounded by the curve π¦ is equal to three π to the power of π₯ over seven and the lines π₯ is equal to zero and π₯ is equal to 14 is equal to 21 times π squared minus one.