find the indefinite integral. use c to represent an arbitrary constant. ∫(7x² - 4/3 e^x - 1) dx =

Explanation:

Step1: Apply integral sum - difference rule

$\int(7x^{2}-\frac{4}{3}e^{x}-1)dx=\int7x^{2}dx-\int\frac{4}{3}e^{x}dx-\int1dx$

Step2: Integrate each term

For $\int7x^{2}dx$, use the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), so $\int7x^{2}dx=7\times\frac{x^{2 + 1}}{2+1}=\frac{7}{3}x^{3}$. For $\int\frac{4}{3}e^{x}dx$, since $\int e^{x}dx=e^{x}+C$, then $\int\frac{4}{3}e^{x}dx=\frac{4}{3}e^{x}$. For $\int1dx$, since $\int1dx=x + C$, then $\int1dx=x$.

Step3: Combine the results

$\int(7x^{2}-\frac{4}{3}e^{x}-1)dx=\frac{7}{3}x^{3}-\frac{4}{3}e^{x}-x + C$

Answer:

$\frac{7}{3}x^{3}-\frac{4}{3}e^{x}-x + C$