integral from 0 to 1/10 of xln(10x) — step-by-step solution

⚠ 10x is read as the product 10*x: for a power, write 10^x

⚠ xln(10x) is read as x*ln(10x): use * for multiplication, use ^ for a power

1) Let's compute the integral:

\int(x \cdot \ln (10 x))dx

For

x \cdot \ln (10 x)

, we use integration by parts:

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

, with u1 =

\ln (10 x)

and v2' =

x

2) Let's perform the differentiation u1:

\frac{d}{d x} (\ln (10 x))

Apply the differentiation rule:

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

with f =

10 x

\frac{d}{d x} (10 x) \cdot \frac{1}{10 x}

ℹ

\frac{d}{d x} (10 x) \cdot \frac{1}{10 x}

\frac{\frac{d}{d x} (10 x)}{10 x}

‖

\frac{d}{d x} (10 x)

10

\frac{10}{10 x}

\frac{1}{x}

So,

u1'

corresponds to

\frac{1}{x}

3) Let's compute the integral v2':

\int(x)dx

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

4) Let's calculate

u1' \cdot v2

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

ℹ

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

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

‖

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

\frac{x}{2}

\frac{x}{2}

5) Let's integrate

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

\ln (10 x) \cdot \frac{x^{2}}{2} - \int(\frac{x}{2})dx

ℹ

\ln (10 x) \cdot \frac{x^{2}}{2}

\frac{\ln (10 x) \cdot x^{2}}{2}

‖

- \int(\frac{x}{2})dx

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

\frac{\ln (10 x) \cdot x^{2}}{2} - \frac{1}{2} \cdot \frac{x^{2}}{2}

\frac{\ln (10 x) \cdot x^{2}}{2} - \frac{x^{2}}{4}

\frac{2 \cdot \ln (10 x) \cdot x^{2} - x^{2}}{4}

▷

\frac{x^{2} \cdot (2 \cdot \ln (10 x) - 1)}{4}

The integrand is undefined at

0

: take the antiderivative's limit from inside the interval

⇥5.1) Let's evaluate the limit of

\frac{x^{2} \cdot (2 \cdot \ln (10 x) - 1)}{4}

when

x \to 0

⁺

⇥ The limit of

x^{2}

0

⁺ when

x \to 0

⁺

⇥ The limit of

10 x

is (

10 \cdot 0

)⁺ when

x \to 0

⁺

⇥5.2) Let's evaluate:

10 \cdot 0

⇥

0

⇥ The limit of

\ln (10 x)

-∞

when

x \to 0

⁺

⇥ The limit of

2 \cdot \ln (10 x)

-∞

when

x \to 0

⁺

⇥ The limit of

2 \cdot \ln (10 x) - 1

-∞

when

x \to 0

⁺

⇥For

x^{2} \cdot (2 \cdot \ln (10 x) - 1)

, we encounter the indeterminate form

0 \cdot -∞

x

tends to

0

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

x^{2} \cdot (2 \cdot \ln (10 x) - 1)

as the quotient

\frac{2 \cdot \ln (10 x) - 1}{\frac{1}{x^{2}}}

⇥ The limit of

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

+∞

when

x \to 0

⁺

⇥For

\frac{2 \cdot \ln (10 x) - 1}{\frac{1}{x^{2}}}

, the limit takes the indeterminate form

- \frac{+∞}{+∞}

when

x

→

0

⇥5.4) 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} (2 \cdot \ln (10 x) - 1)

⇥

\frac{d}{d x} (2 \cdot \ln (10 x)) + 0

⇥Apply the differentiation rule:

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

with f =

10 x

⇥

2 \cdot \frac{d}{d x} (10 x) \cdot \frac{1}{10 x}

⇥

\frac{2 \cdot \frac{d}{d x} (10 x)}{10 x}

⇥

\frac{\frac{d}{d x} (10 x)}{5 x}

⇥

\frac{10}{5 x}

⇥

\frac{2}{x}

⇥5.5) Let's differentiate:

\frac{d}{d x} (\frac{1}{x^{2}})

⇥We differentiate

u1

1

and

v2

x^{2}

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}}

⇥5.6) Let's differentiate

u1

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

⇥

0

⇥5.7) Let's differentiate

v2

\frac{d}{d x} (x^{2})

⇥

2 x

⇥5.8) You can evaluate the resulted derivative (

\frac{u1}{v2}

)':

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

⇥ℹ

0 \cdot x^{2}

0

‖

{(x^{2})}^{2}

x^{(2 \cdot 2)}

⇥

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

⇥

- \frac{2 x}{x^{4}}

⇥

- \frac{2}{x^{3}}

⇥With the derivatives of the numerator and the denominator, we obtain a new fraction whose limit can be determined:

⇥

\frac{\frac{2}{x}}{- \frac{2}{x^{3}}}

⇥

- \frac{\frac{2}{x}}{\frac{2}{x^{3}}}

⇥

- \frac{2}{x} \cdot \frac{x^{3}}{2}

⇥ℹ

- \frac{2}{x} \cdot \frac{x^{3}}{2}

- \frac{2 x^{3}}{x \cdot 2}

‖

- \frac{2 x^{3}}{x \cdot 2}

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

⇥

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

⇥

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

⇥

- x^{2}

⇥ The limit of

- x^{2}

0

⁻ when

x \to 0

⁺

⇥ The limit of

\frac{2 \cdot \ln (10 x) - 1}{\frac{1}{x^{2}}}

0

⁻ when

x \to 0

⁺

⇥The limit of

x^{2} \cdot (2 \cdot \ln (10 x) - 1)

0

⁻ when

x \to 0

⁺

⇥So the limit of

\frac{x^{2} \cdot (2 \cdot \ln (10 x) - 1)}{4}

0

when

x \to 0

⁺

6) Let's evaluate over the interval

[0, \frac{1}{10}]

\frac{{(\frac{1}{10})}^{2} \cdot (2 \cdot \ln (10 \cdot \frac{1}{10}) - 1)}{4}

ℹ

10 \cdot \frac{1}{10}

\frac{10}{10}

‖

\frac{10}{10}

1

\frac{{(\frac{1}{10})}^{2} \cdot (2 \cdot \ln (1) - 1)}{4}

\frac{{(\frac{1}{10})}^{2} \cdot (2 \cdot 0 - 1)}{4}

\frac{{(\frac{1}{10})}^{2} \cdot (0 - 1)}{4}

- \frac{{(\frac{1}{10})}^{2}}{4}

- \frac{\frac{1}{10^{2}}}{4}

- \frac{\frac{1}{100}}{4}

- \frac{1}{100} \cdot \frac{1}{4}

▶Integral:

- \frac{1}{400}

7) The numeric value of:

- \frac{1}{400}

equals:

▷

-0.0025