Question

In Exercises \(27-40\) , evaluate each integral using Part 2 of the Fundamental Theorem. Support your answer with NINT if you are unsure. $$\int_{0}^{1}\left(x^{2}+\sqrt{x}\right) d x$$

Short answer

The value of the integral \(\int_{0}^{1}\left(x^{2}+\sqrt{x}\right) d x\) is \(1\).

Step-by-step solution

Determine the Antiderivatives

To evaluate the integral, first find the antiderivative of each term. For \(x^{2}\), the antiderivative is \(\frac{x^{3}}{3}\), by applying the power rule of integration (which states that the integral of \(x^n\) is \(\frac{x^{n+1}}{n+1}\)). For \(\sqrt{x}\), write it as \(x^{1/2}\), and then apply the power rule to find its antiderivative is \(\frac{2x^{3/2}}{3}\). The antiderivative of the entire function is therefore \(\frac{x^{3}}{3} + \frac{2x^{3/2}}{3}\).

Apply the Fundamental Theorem of Calculus

To evaluate the definite integral from \(0\) to \(1\), plug these values into the antiderivative. That is, subtract the value of the antiderivative at \(0\) (the lower limit of integration) from the value of the antiderivative at \(1\) (the upper limit of integration). This results in \(\left(\frac{1^{3}}{3} + \frac{2(1^{3/2})}{3}\right) - \left(\frac{0^{3}}{3} + \frac{2(0^{3/2})}{3}\right)\) = \(\frac{1}{3} + \frac{2}{3} - 0 = 1\).

Write the Final Answer

Write your final answer. The value of the integral \(\int_{0}^{1}\left(x^{2}+\sqrt{x}\right) d x\) is \(1\).