Question

find the exact value.\\(\int_{0}^{1} (x^2 + \sqrt{x}) \\, dx = ?\\)

Explanation:

Step1: Integrate term by term

The integral of \(x^2\) is \(\frac{x^3}{3}\) and the integral of \(\sqrt{x}=x^{\frac{1}{2}}\) is \(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}=\frac{2}{3}x^{\frac{3}{2}}\). So the antiderivative \(F(x)\) of \(x^2 + \sqrt{x}\) is \(\frac{x^3}{3}+\frac{2}{3}x^{\frac{3}{2}}\).

Step2: Apply Fundamental Theorem of Calculus

Evaluate \(F(1)-F(0)\). For \(F(1)\): \(\frac{1^3}{3}+\frac{2}{3}(1)^{\frac{3}{2}}=\frac{1}{3}+\frac{2}{3}\). For \(F(0)\): \(\frac{0^3}{3}+\frac{2}{3}(0)^{\frac{3}{2}} = 0\). Then \(F(1)-F(0)=\frac{1 + 2}{3}=1\).

Answer:

\(1\)