Question

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. (Round your answer to four decimal places.) enclosed by y = In(x), y = 2 - In(x), and x = 4

Answer 1

The area enclosed by the curves
`$y=\ln (x), y=1+\ln (x)$` is approximately 3 square units.

To find the area enclosed by , and x=4, we first need to determine the intersection point of the curves
`$\ln (x)$` and
`$1+\ln (x)$`.

Setting
`$y=\ln (x)$` and
`$y=1+\ln (x)$` equal to each other:

`\ln (x)=1+\ln (x)`

`\ln (x)-\ln (x)=1`

0=1 - This implies there's no intersection.

Therefore, the region enclosed is bounded by
`$y=\ln (x), y=1+\ln (x)$`, and x=4 from x=1 to x=4.

Using integration:
`$\int_1^4((1+\ln (x))-\ln (x)) d x$`

Solving the integral:
`$\int_1^4(1 d x)=\left.x\right|_1 ^4=4-1=3$`

Hence, the area of the enclosed region is 3 square units.

Question:-

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. (Round your answer to four decimal places.)

enclosed by y = In(x), y=1In(x), and x = 4