integ(exp(-x), x, [0, +inf])

1) Before integrating, let's calculate:

\exp (- x)

ⓘ The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base:

\exp (- x)

\frac{1}{\exp (x)}

\frac{1}{\exp (x)}

2) Let's integrate:

\int_{0}^{+∞} \frac{1}{\exp (x)} d x

For the math expression:

\frac{1}{\exp (x)}

, knowing that

{(\frac{1}{a^{n}})}^{'} = - \frac{{(n \cdot \ln (a))}^{'}}{a^{n}}

, let's try differentiating

\frac{1}{\exp (x)}

3) Let's perform the differentiation:

\frac{d}{d x} (\frac{1}{\exp (x)})

We differentiate

p1

1

and

q2

\exp (x)

using:

\frac{d}{d x} (\frac{p1}{q2}) = \frac{\frac{d}{d x} (p1) \cdot q2 - p1 \cdot \frac{d}{d x} (q2)}{{q2}^{2}}

4) Let's perform the differentiation

p1

\frac{d}{d x} (1)

ⓘ The derivative of a constant is zero:

\frac{d}{d x} (1)

0

0

5) Let's perform the differentiation

q2

\frac{d}{d x} (\exp (x))

ⓘ The derivative of

\exp (x)

\exp (x)

\frac{d}{d x} (\exp (x))

\exp (x)

\exp (x)

6) We evaluate the resulting derivative (

\frac{p1}{q2}

)':

\frac{0 \cdot \exp (x) - 1 \cdot \exp (x)}{{(\exp (x))}^{2}}

ⓘ A product with a zero factor is zero:

0 \cdot \exp (x)

0

■A power of a power multiplies the exponents,

{(a^{n})}^{m} = a^{n m}

{(\exp (x))}^{2}

\exp (x \cdot 2)

\frac{0 - \exp (x)}{\exp (x \cdot 2)}

ⓘ Adding 0 preserves the original value:

0 - \exp (x)

- \exp (x)

■The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor:

- \frac{\exp (x)}{\exp (x \cdot 2)}

- \exp (x - x \cdot 2)

- \exp (x - x \cdot 2)

ⓘ Group like terms by adding their coefficients:

x - x \cdot 2

- x

- \exp (- x)

ⓘ The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base:

- \exp (- x)

- \frac{1}{\exp (x)}

- \frac{1}{\exp (x)}

7) Then you can resume the derivation:

ⅰ

\frac{d}{d x} (\frac{1}{\exp (x)})

- \frac{1}{\exp (x)}

- \frac{1}{\exp (x)}

You observe that the computed derivative and the original math expression are the same, except for a coefficient. So the integral is:

ⓘ Differentiating

- \frac{1}{\exp (x)}

gives

\frac{1}{\exp (x)}

, so

- \frac{1}{\exp (x)}

is an antiderivative of

\frac{1}{\exp (x)}

\int \frac{1}{\exp (x)} d x

- \frac{1}{\exp (x)}

▷

- \frac{1}{\exp (x)}

At the infinite bound

+∞

, the antiderivative is evaluated by its limit

Let's evaluate the limit of

- \frac{1}{\exp (x)}

when

x \to +∞

ⓘ This is a reference limit of the function exp

The limit of

\exp (x)

+∞

when

x \to +∞

ⓘ A number divided by a quantity that grows without bound tends to 0

So the limit of

- \frac{1}{\exp (x)}

0

⁻ when

x \to +∞

8) Let's evaluate for the interval

[0, +∞]

\frac{1}{\exp (0)}

ⓘ Any nonzero number raised to the power 0 is 1:

\exp (0)

1

▶Integral:

1

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