$\int_0^y e^x dx = e$, solve for y

Question

$\int_0^y e^x dx = e$

What is $y$ here? The definite integral from $0$ to $1$ is $e - 1$. But what number must be this integral's upper limit in order to produce an area under the curve of $e$?

Approximating with a graphing calculator, I seem to get $1.313261687$. Is this correct?

Answer 1

$$\int_0^{y} e^x dx = \left|e^x\right|_{0}^{y}$$ $$\rightarrow e^y - e^0 = e^y - 1$$

It is given that this expression equals $e$. So lets solve it from here:

$$e^y - 1 = e$$ $$e^y = e + 1$$

Taking natural log on both sides we get

$$y = ln(e + 1)$$

and using a calculator you can verify that $ln(e + 1) = 1.313261...$

Answer 2

Here $e^y - e^0 = e$, then $e^y = e+1$ and so $y = \ln(e+1)$.

Answer 3

$\int_0^y e^x dx = e^y-1 = e \Rightarrow e^y = 1+e$, which means $y = \ln (1+e) \approx 1.3132616875182228 $

Answer 4

We can solve the question by finding its antiderivative $\int_0^y e^x dx = e]^y_0= e^y - e^0 = e^y - 1$

Solving for $e^y - 1 = e$

$e^y - 1 = e$
$e^y = e + 1$
$y = ln(e+1)$

So, your approximation is right. But ln(e+1) is the exact solution.