Evaluate the definite integral of sqrt(x - x^4) with respect to x, from x = 0 to x = 1.

Step-by-step solution

1. 1

   Factor inside the radical. Writing $x-x^{4}=x\left(1-x^{3}\right)$ splits the integrand into two independent powers:

   $I=\int_{0}^{1}x^{1/2}\left(1-x^{3}\right)^{1/2}dx.$

   On $[0,1]$ both factors are non-negative, so the square root is real throughout and the integral is proper. The shape $x^{p}(1-x^{q})^{r}$ is the signature of a Beta-function integral.

2. 2

   Substitute $u=x^{3}$ to match the Beta form. Then $x=u^{1/3}$, $du=3x^{2}dx$ so $dx=\tfrac13u^{-2/3}du$, and $x^{1/2}=u^{1/6}$. The limits are unchanged, $0\to0$ and $1\to1$:

   $I=\int_{0}^{1}u^{1/6}(1-u)^{1/2}\cdot\tfrac13u^{-2/3}\,du.$

3. 3

   Combine the powers of $u$. Adding the exponents,

   $\frac16-\frac23=\frac16-\frac46=-\frac12,$

   so

   $I=\frac13\int_{0}^{1}u^{-1/2}(1-u)^{1/2}\,du.$

   This is exactly the Beta integral $B(a,b)=\int_{0}^{1}u^{a-1}(1-u)^{b-1}du$ with $a-1=-\tfrac12$ and $b-1=\tfrac12$, i.e. $a=\tfrac12$, $b=\tfrac32$.

4. 4

   Evaluate with Gamma functions. Using $B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$ together with $\Gamma\!\left(\tfrac12\right)=\sqrt{\pi}$, $\Gamma\!\left(\tfrac32\right)=\tfrac{\sqrt{\pi}}{2}$ and $\Gamma(2)=1$:

   $I=\frac13\cdot\frac{\sqrt{\pi}\cdot\frac{\sqrt{\pi}}{2}}{1}=\frac13\cdot\frac{\pi}{2}=\frac{\pi}{6}.$

5. 5

   Confirm numerically. Composite Simpson quadrature on $[0,1]$ with two million subintervals gives $0.52359878$, against $\frac{\pi}{6}=0.52359878$ ✓. The agreement is slightly limited by the square-root singularity in the derivative at both endpoints, which is exactly why the Beta-function route is preferable to numerical integration here.