Question

Compute the area of the curvilinear trapezoid bounded by the \(x\)-axis and the curve \(y=x-x^{2} \sqrt{x}\).

Short answer

Answer: The area of the curvilinear trapezoid is \(\frac{3}{7}\).

Step-by-step solution

Determine the range where the curve is above the \(x\)-axis

To find the range where the curve is above the \(x\)-axis, we need to find the points where the curve intersects the \(x\)-axis. In other words, we need to find the values of \(x\) when \(y = 0\) in the equation \(y = x - x^2 \sqrt{x}\). Set \(y=0\) and solve for \(x\): \begin{align*} 0 = x - x^2 \sqrt{x}. \end{align*} Factor out an \(x\): \begin{align*} 0 = x (1 - x \sqrt{x}). \end{align*} There are two solutions: \(x = 0\) and \(1 - x \sqrt{x} = 0\). Solve for the nonzero solution: \begin{align*} 1 - x \sqrt{x} = 0. \end{align*} \(x \sqrt{x} = 1\). Square both sides: \(x^3 = 1\). The only possible solution is \(x=1\). Thus, the range where the curve is above the \(x\)-axis is \(0 \leq x \leq 1\).

Integrate the curve over the range

Now that we have the range, we can integrate the curve \(y = x - x^2 \sqrt{x}\) over the range \(0 \leq x \leq 1\) to find the area of the curvilinear trapezoid. \begin{align*} A = \int_{0}^{1} (x - x^2 \sqrt{x}) dx. \end{align*} To make integration easier, let's rewrite the equation: \begin{align*} A = \int_{0}^{1} (x - x^{5/2}) dx. \end{align*} Now, integrate each term: \begin{align*} A = \left[ \frac{1}{2}x^2 - \frac{2}{7}x^{7/2} \right]_{0}^{1}. \end{align*}

Evaluate the integral

We now evaluate the integral at the boundaries to find the area: \begin{align*} A &= \left[ \frac{1}{2}(1)^2 - \frac{2}{7}(1)^{7/2} \right] - \left[ \frac{1}{2}(0)^2 - \frac{2}{7}(0)^{7/2} \right] \\ &= \left[ \frac{1}{2} - \frac{2}{7} \right] \\ &= \frac{3}{7}. \end{align*} The area of the curvilinear trapezoid bounded by the \(x\)-axis and the curve \(y = x - x^2 \sqrt{x}\) is \(\frac{3}{7}\).