Question

evaluate the integral.
\\(\int_{1}^{4} (2x^2 + \sqrt{x}) dx\\)
\\(\int_{1}^{4} (2x^2 + \sqrt{x}) dx = \square\\) (simplify your answer.)

Explanation:

Step1: Integrate term by term

First, rewrite \(\sqrt{x}\) as \(x^{\frac{1}{2}}\). Then integrate \(2x^{2}\) and \(x^{\frac{1}{2}}\) separately. The integral of \(ax^{n}\) is \(\frac{a}{n + 1}x^{n+1}+C\) (we can ignore the constant \(C\) for definite integrals).

For \(2x^{2}\): \(\int 2x^{2}dx=\frac{2}{3}x^{3}+C\)

For \(x^{\frac{1}{2}}\): \(\int x^{\frac{1}{2}}dx=\frac{2}{3}x^{\frac{3}{2}}+C\)

So the antiderivative of \(2x^{2}+\sqrt{x}\) is \(\frac{2}{3}x^{3}+\frac{2}{3}x^{\frac{3}{2}}\)

Step2: Apply the Fundamental Theorem of Calculus

Evaluate the antiderivative at the upper limit (4) and subtract the value at the lower limit (1).

First, at \(x = 4\):

\(\frac{2}{3}(4)^{3}+\frac{2}{3}(4)^{\frac{3}{2}}=\frac{2}{3}(64)+\frac{2}{3}(8)=\frac{128}{3}+\frac{16}{3}=\frac{144}{3} = 48\)

Then, at \(x = 1\):

\(\frac{2}{3}(1)^{3}+\frac{2}{3}(1)^{\frac{3}{2}}=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}\)

Now subtract: \(48-\frac{4}{3}=\frac{144 - 4}{3}=\frac{140}{3}\)

Answer:

\(\frac{140}{3}\)