Question

$\int_{3}^{4} x^2 + 4x \\, dx$

Explanation:

Step1: Find antiderivative

The antiderivative of \(x^2\) is \(\frac{1}{3}x^3\) and the antiderivative of \(4x\) is \(2x^2\). So the antiderivative of \(x^2 + 4x\) is \(F(x)=\frac{1}{3}x^3+2x^2\).

Step2: Apply Fundamental Theorem

Evaluate \(F(4)-F(3)\). First, \(F(4)=\frac{1}{3}(4)^3 + 2(4)^2=\frac{64}{3}+32=\frac{64 + 96}{3}=\frac{160}{3}\). Then, \(F(3)=\frac{1}{3}(3)^3+2(3)^2 = 9 + 18 = 27=\frac{81}{3}\). Now, \(F(4)-F(3)=\frac{160}{3}-\frac{81}{3}=\frac{79}{3}\).

Answer:

\(\frac{79}{3}\)