evaluate the definite integral below. ∫₁³ (7x² - 3) dx

Explanation:

Step1: Find antiderivative

The antiderivative of $7x^{2}-3$ is $\frac{7}{3}x^{3}-3x$ using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$).

Step2: Apply fundamental theorem

$\left(\frac{7}{3}(3)^{3}-3(3)\right)-\left(\frac{7}{3}(1)^{3}-3(1)\right)$. First, calculate $\frac{7}{3}(3)^{3}-3(3)$: $\frac{7}{3}(3)^{3}-3(3)=\frac{7}{3}\times27 - 9=63 - 9 = 54$. Then, calculate $\frac{7}{3}(1)^{3}-3(1)$: $\frac{7}{3}(1)^{3}-3(1)=\frac{7}{3}-3=\frac{7 - 9}{3}=-\frac{2}{3}$. Finally, subtract: $54-\left(-\frac{2}{3}\right)=54+\frac{2}{3}=\frac{162 + 2}{3}=\frac{164}{3}\approx54.67$.

Answer:

$\frac{164}{3}$