Rregion in the first quadrant bounded by the axes, $y = e^x$, $x = e^y$, and the line $x = 4$.

Question

What integral gives the area of the region in the first quadrant bounded by the axes, $y = e^x$, $x = e^y$, and the line $x = 4$.

The integral from $0$ to $4$ of the difference between $e$ to the $x$ power and the natural $\log$ of $x$, $dx$.

that was what i got but im not sure if that is correct. Please help

Answer 1

You can (or maybe have to) get it from two integrals (sketch the graphs!):

$$\int_0^1e^xdx+\int_1^4\left(e^x-\log x\right)dx$$

Answer 2

Note that for $x<1$, $\log(x)<0$. Therefore, for $0\le x\le 1$, the area is the area under the curve $e^x$. For $1\le x\le 4$, $e^x>\log(x)\ge 0$.

Putting it together, we can write

$$\text{Area}=\int_0^1 e^x\,dx+\int_1^4 (e^x-\log(x))\,dx$$

Answer 3

Draw graphic you can see that $$\int _{ 0 }^{ 1 }{ { e }^{ x }dx+\int _{ 1 }^{ 4 }{ \left[ { e }^{ x }-\ln { x } \right] dx } } $$