Integral of x⁵ ln(x + 1) from 0 to 2

Grades 11-12

Question: 27. y = x^5 ln(x + 1), 0 ≤ x ≤ 2

The function

y = x^5 \ln(x + 1)

is a product of two distinct types of functions: a high-degree polynomial (

x^5

) and a logarithmic function (

\ln(x + 1)

). In the interval

[0, 2]

, we can observe several interesting mathematical properties:

x = 0

, the function evaluates to

0^5 \cdot \ln(1) = 0 \cdot 0 = 0

. Because of the

x^5

factor, the graph is extremely "flat" near the origin. The first four derivatives of

x^5

are all zero at

x = 0

, which means the curve hugs the x-axis very closely before it starts to rise.

x

increases from 0 toward 2, the function grows very rapidly.

The $x^5$ term contributes a massive scaling factor (at $x=2$ , $x^5 = 32$ ).
The $\ln(x+1)$ term grows much more slowly (at $x=2$ , $\ln(3) \approx 1.098$ ).
The product results in a value of approximately $35.15$ at the endpoint $x = 2$ .

The slope of the curve is found using the product rule:

\begin{align*} \frac{dy}{dx} &= \frac{d}{dx}(x^5) \cdot \ln(x + 1) + x^5 \cdot \frac{d}{dx}(\ln(x + 1)) \\ \frac{dy}{dx} &= 5x^4 \ln(x + 1) + \frac{x^5}{x + 1} \end{align*}

Since both

x^4 \ln(x+1)

and

\frac{x^5}{x+1}

are positive for all

x > 0

, the function is strictly increasing on the given interval.

The Curve (Blue): Represents the function $f(x) = x^5 \ln(x + 1)$ . Notice how it stays near zero for a while and then shoots upward.
Point P: You can move the slider to trace the point along the curve from $x=0$ to $x=2$ .
Tangent Line (Red): Shows the instantaneous rate of change at any point. Notice how the tangent line becomes nearly vertical as $x$ approaches 2, reflecting the extremely high slope ( $5 \cdot 16 \ln(3) + 32/3 \approx 52.7 + 10.7 \approx 63.4$ ).
Bounding Box: The graph is focused on the domain $[0, 2]$ and the corresponding range $[0, 35.15]$ .