Definite Integral of x⁴/(x + 1) from 0 to 4

Grades 11-12

Question: IntegralEntre(x^4/(x+1),4)

To calculate the integral of the rational function

\frac{x^4}{x+1}

between the bounds (typically from

0

4

as suggested by the problem), we use a technique called polynomial long division followed by term-by-term integration.

Since the degree of the numerator (

4

) is greater than the degree of the denominator (

1

), we first simplify the expression:

\begin{align*} \frac{x^4}{x+1} &= \frac{x^4 + x^3 - x^3}{x+1} \\ &= x^3 - \frac{x^3}{x+1} \\ &= x^3 - \frac{x^3 + x^2 - x^2}{x+1} \\ &= x^3 - x^2 + \frac{x^2}{x+1} \\ &= x^3 - x^2 + x - \frac{x}{x+1} \\ &= x^3 - x^2 + x - 1 + \frac{1}{x+1} \end{align*}

Now we integrate each term individually:

\begin{align*} \int \left( x^3 - x^2 + x - 1 + \frac{1}{x+1} \right) dx &= \frac{x^4}{4} - \frac{x^3}{3} + \frac{x^2}{2} - x + \ln|x+1| + C \end{align*}

To find the area under the curve from

x=0

x=4

\begin{align*} \left[ \frac{x^4}{4} - \frac{x^3}{3} + \frac{x^2}{2} - x + \ln(x+1) \right]_0^4 &= \left( \frac{256}{4} - \frac{64}{3} + \frac{16}{2} - 4 + \ln(5) \right) - (0 - 0 + 0 - 0 + \ln(1)) \\ &= \left( 64 - 21.333 + 8 - 4 + 1.609 \right) \\ &\approx 48.276 \end{align*}

In the accompanying diagram:

The blue curve represents the function $f(x) = \frac{x^4}{x+1}$ .
The shaded region represents the accumulated area from $x=0$ to the limit $b$ (set by the slider).
You can observe how the vertical asymptote at $x = -1$ affects the shape of the graph, though we are integrating in the positive region.