integral from 0 to 1 of ln(4x) — step-by-step solution

1) Let's compute the integral:

\int(\ln (4 x))dx

For

\ln (4 x)

, we use integration by parts:

\int(u1 \cdot v2')dx = u1 \cdot v2 - \int(u1' \cdot v2)dx

, with u1 =

\ln (4 x)

and v2' =

1

2) Let's perform the differentiation u1:

\frac{d}{d x} (\ln (4 x))

We differentiate using:

\frac{d}{d x} (\ln (f)) = \frac{d}{d x} (f) \cdot \frac{1}{f}

(where f =

4 x

)

\frac{d}{d x} (4 x) \cdot \frac{1}{4 x}

ℹ

\frac{d}{d x} (4 x) \cdot \frac{1}{4 x}

\frac{\frac{d}{d x} (4 x)}{4 x}

‖

\frac{d}{d x} (4 x)

4

\frac{4}{4 x}

\frac{1}{x}

u1'

is equal to

\frac{1}{x}

3) Let's compute the integral v2':

\int(1)dx

x

4) Let's calculate

u1' \cdot v2

\frac{1}{x} \cdot x

ℹ

\frac{1}{x} \cdot x

\frac{x}{x}

‖

\frac{x}{x}

1

1

5) Let's compute the integral

u1 \cdot v2 - \int(u1' \cdot v2)dx

\ln (4 x) \cdot x - \int(1)dx

\ln (4 x) \cdot x - x

The integrand is undefined at

0

: take the antiderivative's limit from inside the interval

⇥5.1) Compute the limit of

x \cdot \ln (4 x) - x

x \to 0

⁺

⇥ The limit of

4 x

4 \cdot 0

⁺ when

x \to 0

⁺

⇥⇥5.1.1) Let's calculate:

4 \cdot 0

⇥⇥

0

⇥ The limit of

\ln (4 x)

-∞

when

x \to 0

⁺

⇥For

x \cdot \ln (4 x)

, we encounter the indeterminate form

0 \cdot -∞

x

tends to

0

⇥5.2) To resolve the indeterminate form 0 × ∞, we rewrite the product

x \cdot \ln (4 x)

as the quotient

\frac{\ln (4 x)}{\frac{1}{x}}

⇥ The limit of

\frac{1}{x}

+∞

when

x \to 0

⁺

⇥For

\frac{\ln (4 x)}{\frac{1}{x}}

, the limit takes the indeterminate form

- \frac{+∞}{+∞}

when

x

→

0

⇥5.3) Let's apply L'Hôpital's rule (limit of a/b = limit of a'/b') by differentiating numerator and denominator:

⇥Let's differentiate:

\frac{d}{d x} (\ln (4 x))

⇥We differentiate using:

\frac{d}{d x} (\ln (f)) = \frac{d}{d x} (f) \cdot \frac{1}{f}

(where f =

4 x

)

⇥

\frac{d}{d x} (4 x) \cdot \frac{1}{4 x}

⇥ℹ

\frac{d}{d x} (4 x) \cdot \frac{1}{4 x}

\frac{\frac{d}{d x} (4 x)}{4 x}

‖

\frac{d}{d x} (4 x)

4

⇥

\frac{4}{4 x}

⇥

\frac{1}{x}

⇥5.4) Let's perform the differentiation:

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

⇥We differentiate

u1

1

and

v2

x

using:

\frac{d}{d x} (\frac{u1}{v2}) = \frac{\frac{d}{d x} (u1) \cdot v2 - u1 \cdot \frac{d}{d x} (v2)}{{v2}^{2}}

⇥5.5) Let's perform the differentiation

u1

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

⇥

0

⇥5.6) Let's perform the differentiation

v2

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

⇥

1

⇥5.7) Let's evaluate the resulting derivative (

\frac{u1}{v2}

)':

\frac{0 \cdot x - 1 \cdot 1}{x^{2}}

⇥ℹ

0 \cdot x

0

‖

- 1 \cdot 1

-1

⇥

\frac{0 - 1}{x^{2}}

⇥

- \frac{1}{x^{2}}

⇥Using those derivatives, we obtain a new fraction whose limit can be computed:

⇥

\frac{\frac{1}{x}}{- \frac{1}{x^{2}}}

⇥

- \frac{\frac{1}{x}}{\frac{1}{x^{2}}}

⇥

- \frac{1}{x} \cdot x^{2}

⇥ℹ

- \frac{1}{x} \cdot x^{2}

- \frac{x^{2}}{x}

‖

- \frac{x^{2}}{x}

- x

⇥

- x

⇥ The limit of

\frac{\ln (4 x)}{\frac{1}{x}}

0

⁻ when

x \to 0

⁺

⇥The limit of

x \cdot \ln (4 x)

0

⁻ when

x \to 0

⁺

⇥So the limit of

x \cdot \ln (4 x) - x

0

⁻ when

x \to 0

⁺

6) Let's evaluate over the interval

[0, 1]

1 \cdot \ln (4 \cdot 1) - 1

\ln (4) - 1

▶Integral:

2 \cdot \ln (2) - 1

7) The numeric value of:

2 \cdot \ln (2) - 1

is:

2 \cdot \approx 0.6931 - 1

\approx 1.3863 - 1

▷

\approx 0.3863