Substitution leads me to wrong result for the integral $\int_0^x\lambda e^{-\lambda x}dx$

Question

I have a question that should have been trivial to solve, but for some reason I keep messing up. I need to calculate the following integral:

$$\int_0^x\lambda e^{-\lambda x}dx$$

I initially tried to solve this with the substitution:

$$u=-\lambda x$$

This leads to the following:

$$ \begin{aligned} x&=-\frac{u}{\lambda}\\ dx&=-\frac{du}{\lambda} \end{aligned}$$

My first (wrong) attempt:

$$ \begin{aligned} \int_0^x\lambda e^{-\lambda x}dx &= \lambda\int_0^{\frac{-u}{\lambda}} e^{u}\frac{-du}{\lambda}\\ & = -\int_0^{\frac{-u}{\lambda}} e^{u}du = -[e^u]_0^{\frac{-u}{\lambda}}\\ &= -[e^{\frac{-u}{\lambda}}-e^0] = 1 - e^{\frac{-u}{\lambda}} = 1 - e^x \end{aligned} $$

Online tools have shown that the answer is supposed to be $1 - e^{-\lambda x}$ however.

My second slightly modified attempt:

I reuse the steps from my first attempt all the way until I get:

$$-[e^u]_0^{\frac{-u}{\lambda}}$$

and now I replace the substitution first before continuing:

$$ \begin{aligned} -[e^u]_0^{\frac{-u}{\lambda}} &= -[e^{-\lambda x}]_0^x\\ & = -[e^{-\lambda x} - e^0]\\ & = 1 - e^{-\lambda x} \end{aligned} $$

What sorcery is this? I can't seem to recall a rule about substituting the variables back to the original before filling in the limits. Otherwise, what is the point of changing the limits along with the substitution? I must be missing something obvious here, but not obvious enough for me to notice.

Without substitution works without a problem:

$$ \begin{aligned} \int_0^x\lambda e^{-\lambda x}dx &= -[e^{-\lambda x}]_0^xdx\\ & = -[e^{-\lambda x} - 1]\\ &= 1 - e^{-\lambda x} \end{aligned} $$

Why was my first attempt wrong?

Answer 1

In your wrong attempt you wrote: $$\int_0^x\lambda e^{-\lambda x} \ \mathrm dx = \lambda\int_0^{\color{red}{\frac{-u}{\lambda}}} e^{u}\frac{-\mathrm du}{\lambda}$$ when in fact you should have written: $$\int_0^x\lambda e^{-\lambda x} \ \mathrm dx = \lambda\int_0^{\color{red}u} e^{u}\frac{-\mathrm du}{\lambda}$$

When changing the bounds, ask yourself "when $x$ = lower bound, what is $u$? when $x$ = upper bound, what is $u$?"

The answer is "when $x=0$, $u=0$; when $x=x$, $u=u$".

Answer 2

One mistake at the start with the substitution $u=-\lambda x$ you should get $$dx = -\frac{du}{\lambda} $$ but for some reason you get $$dx = -\frac{du}{x}$$

If you keep going from that spot (and change integral limit to for example $X$ as proposed by Nicolas Francois) you should be able to solve it.