area 6ln(x),xln(x) — step-by-step solution

⚠Ⓘ 6ln is read as the product 6*ln(...): for a power, write 6^ln(...)‖(6ln(x),xln(x)) is read as (6ln(x),x*ln(x)): use * for multiplication, use ^ for a power

y

is already isolated in

y = 6 \cdot \ln (x)

y

is already isolated in

y = x \cdot \ln (x)

At the intersections,

6 \cdot \ln (x)

x \cdot \ln (x)

. Subtract the second expression from the first:

6 \cdot \ln (x) - x \cdot \ln (x) = 0

⇥You can factor:

6 \cdot \ln (x) - x \cdot \ln (x)

into:

\ln (x) \cdot (6 - x)

⇥1.1) Let's solve in

ℝ

\ln (x) = 0

⇥Let's apply the exp function on both sides:

⇥

e^{\ln (x)} = e^{0}

⇥Solution obtained:

x = 1

⇥1.2) Let's solve in

ℝ

6 - x = 0

⇥Solution found:

x = 6

⇥1.3) Let's verify the equation for x:

x > 0

where

x = 1

⇥

1 > 0

⇥The solved variables are indeed compatible with this equation.

⇥1.4) Let's check the equation for x:

x > 0

where

x = 6

⇥

6 > 0

⇥The solved variables are compatible with this equation.

The area between

6 \cdot \ln (x)

and

x \cdot \ln (x)

[1, 6]

is the integral of the absolute value of their difference

No interior intersection: the difference keeps its sign on the interval

[1, 6]

6 \cdot \ln (x)

is above

x \cdot \ln (x)

: integrate

6 \cdot \ln (x) - x \cdot \ln (x)

⇥1.5) Let's compute the integral:

\int(6 \cdot \ln (x) - x \cdot \ln (x))dx

⇥

\int(6 \cdot \ln (x))dx - \int(x \cdot \ln (x))dx

For

6 \cdot \ln (x)

, we use integration by parts:

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

, with u1 =

\ln (x)

and v2' =

1

2) Let's perform the differentiation u1:

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

\frac{1}{x}

3) Let's integrate v2':

\int(1)dx

x

4) Let's evaluate

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 integrate

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

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

ℹ

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

6 \cdot \ln (x) \cdot x + 6 \cdot (- \int(1)dx)

‖

- \int(1)dx

- x

6 \cdot \ln (x) \cdot x + 6 \cdot (- x)

6 \cdot \ln (x) \cdot x - 6 x

⇥Then you can continue the integration:

For

x \cdot \ln (x)

, we use integration by parts:

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

, with u1 =

\ln (x)

and v2' =

x

6) Let's perform the differentiation u1:

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

\frac{1}{x}

7) Let's compute the integral v2':

\int(x)dx

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

8) Let's evaluate

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}

9) Let's compute the integral

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

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

ℹ

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

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

‖

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

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

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

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

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

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

⇥Finally you can finish the integral computation:

⇥

(6 \cdot \ln (x) \cdot x - 6 x) - \frac{x^{2} \cdot (2 \cdot \ln (x) - 1)}{4}

▷

6 \cdot \ln (x) \cdot x - 6 x - \frac{x^{2} \cdot (2 \cdot \ln (x) - 1)}{4}

⇥9.1) Let's evaluate over the interval

[1, 6]

(- \frac{6^{2} \cdot (2 \cdot \ln (6) - 1)}{4} + 6 \cdot 6 \cdot \ln (6) - 6 \cdot 6) - (- \frac{1^{2} \cdot (2 \cdot \ln (1) - 1)}{4} + 6 \cdot 1 \cdot \ln (1) - 6 \cdot 1)

⇥ℹ

6^{2}

36

‖

6 \cdot 6

36

‖

- 6 \cdot 6

-36

‖

- (- \frac{1^{2} \cdot (2 \cdot \ln (1) - 1)}{4} + 6 \cdot 1 \cdot \ln (1) - 6 \cdot 1)

\frac{1^{2} \cdot (2 \cdot \ln (1) - 1)}{4} - 6 \cdot 1 \cdot \ln (1) + 6 \cdot 1

⇥

- \frac{36 \cdot (2 \cdot \ln (6) - 1)}{4}

+ 36 \cdot \ln (6)

- 36

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

- 6 \cdot 1 \cdot \ln (1)

+ 6 \cdot 1

⇥

- 9 \cdot (2 \cdot \ln (6) - 1)

+ 36 \cdot \ln (6)

- 36

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

- 6 \cdot \ln (1)

+ 6

⇥ℹ

- 36 + 6

-30

‖

\frac{2 \cdot \ln (1) - 1}{4} - 6 \cdot \ln (1)

\frac{(2 \cdot \ln (1) - 1) + 4 \cdot (- 6 \cdot \ln (1))}{4}

⇥

- 9 \cdot (2 \cdot \ln (6) - 1)

+ 36 \cdot \ln (6)

- 30

+ \frac{(2 \cdot \ln (1) - 1) + 4 \cdot (- 6 \cdot \ln (1))}{4}

⇥ℹ

\ln (1)

0

‖

4 \cdot (-6)

-24

⇥

- 9 \cdot (2 \cdot \ln (6) - 1)

+ 36 \cdot \ln (6)

- 30

+ \frac{2 \cdot 0 - 1 - 24 \cdot 0}{4}

⇥ℹ

2 \cdot 0

0

‖

-24 \cdot 0

0

⇥

- 9 \cdot (2 \cdot \ln (6) - 1)

+ 36 \cdot \ln (6)

- 30

+ \frac{0 - 1 + 0}{4}

⇥

- 9 \cdot (2 \cdot \ln (6) - 1) + 36 \cdot \ln (6) + \frac{4 \cdot (-30) - 1}{4}

⇥

- 9 \cdot (2 \cdot \ln (6) - 1) + 36 \cdot \ln (6) + \frac{- 120 - 1}{4}

⇥

- 9 \cdot (2 \cdot \ln (6) - 1) + 36 \cdot \ln (6) - \frac{121}{4}

⇥

- (18 \cdot \ln (6) - 9) + 36 \cdot \ln (6) - \frac{121}{4}

⇥

- 18 \cdot \ln (6)

+ 9

+ 36 \cdot \ln (6)

- \frac{121}{4}

⇥ℹ

9 - \frac{121}{4}

\frac{4 \cdot 9 - 121}{4}

‖

- 18 \cdot \ln (6) + 36 \cdot \ln (6)

18 \cdot \ln (6)

⇥

18 \cdot \ln (6) + \frac{4 \cdot 9 - 121}{4}

⇥

18 \cdot \ln (6) + \frac{36 - 121}{4}

⇥

18 \cdot \ln (6) - \frac{85}{4}

The area on

[1, 6]

18 \cdot \ln (6) - \frac{85}{4}

▶Area:

18 \cdot \ln (6) - \frac{85}{4}

10) The numeric value of:

18 \cdot \ln (6) - \frac{85}{4}

equals:

18 \cdot \approx 1.7918 - 21.25

\approx 32.2517 - 21.25

▷

\approx 11.0017