Question

Find the area of the region enclosed by​

Answer 1

7.5 -ln(0.25)

Explanation:

We can first graph this function. Looking at the graph, we can separate this into two ranges -- from x=0.25 to x=1 and from x=1 to x=4.

From x=0.25 to x=1, we want to find the area above y=1/x and below y=4. To do this, we can take the integral of y=4 from x=0.25 to x=1 to encompass the whole area under y=4 from that range. Then, we subtract the area under y=1/x as we don't need that (we only want what's above it). Therefore, we have

`\int\limits^(1)_(0.25){4} \, dx - \int\limits^(1)_(0.25) {1/x} \, dx \\= 4(1) - 4 (0.25) - (ln(1) - ln(0.25))\\= 4 - 1 - ln(1) + ln(0.25)\\= 3 - ln(1) + ln(0.25)`

Next, we know that ln(x/y) = ln(x) - ln(y), so ln(0.25) - ln(1) = ln(0.25/1) = ln(0.25), so we have

3 - ln(0.25)

For x=1 to x=4, we want to find the area above y=x and below y=4. We can apply a similar strategy, subtracting the area below y=x from the area below y=4 to get

`\int\limits^(4)_(1){4} \, dx - \int\limits^(4)_(1) {x} \, dx \\ \\= 4(4) - 4(1) - (4^(2) /2 - 1^2/2)\\= 16 - 4 - 8 + 0.5\\= 4.5`

Add these two areas up to get

3-ln(0.25) + 4.5 = 7.5 -ln(0.25)