Calculus worksheet featuring ten problems focused on applying the power rule to find indefinite integrals.

Here are the solutions for each problem on the worksheet. I have worked through them step-by-step to ensure accuracy.

The Rule Used:
To integrate a power of $x$ (like $x^n$), you add 1 to the exponent and divide by the new exponent:
$$ \int x^n dx = \frac{x^{n+1}}{n+1} + C $$
*(Don't forget to add $+ C$ at the end for indefinite integrals!)*

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1) $\int (2x - 1)dx$
* Rewrite $2x$ as $2x^1$. The integral of $x^1$ is $\frac{x^2}{2}$. So, $2(\frac{x^2}{2}) = x^2$.
* The integral of a constant like $-1$ is just $-1x$ or $-x$.
* Result: $x^2 - x + C$

2) $\int (-4x^3 + 36x^2 - 72x)dx$
* Term 1: $\int -4x^3 dx = -4(\frac{x^4}{4}) = -x^4$
* Term 2: $\int 36x^2 dx = 36(\frac{x^3}{3}) = 12x^3$
* Term 3: $\int -72x dx = -72(\frac{x^2}{2}) = -36x^2$
* Result: $-x^4 + 12x^3 - 36x^2 + C$

3) $\int (2x + 3)dx$
* Term 1: $\int 2x dx = 2(\frac{x^2}{2}) = x^2$
* Term 2: $\int 3 dx = 3x$
* Result: $x^2 + 3x + C$

4) $\int (\frac{-1}{x^2})dx$
* Rewrite using negative exponents: $\int -x^{-2} dx$
* Add 1 to exponent: $-2 + 1 = -1$
* Divide by new exponent: $\frac{-x^{-1}}{-1} = x^{-1}$
* Rewrite as fraction: $\frac{1}{x}$
* Result: $\frac{1}{x} + C$

5) $\int (\frac{-57}{x^4})dx$
* Rewrite: $\int -57x^{-4} dx$
* Add 1 to exponent: $-4 + 1 = -3$
* Divide by new exponent: $\frac{-57x^{-3}}{-3} = 19x^{-3}$
* Rewrite as fraction: $\frac{19}{x^3}$
* Result: $\frac{19}{x^3} + C$

6) $\int (\frac{-28}{x^3})dx$
* Rewrite: $\int -28x^{-3} dx$
* Add 1 to exponent: $-3 + 1 = -2$
* Divide by new exponent: $\frac{-28x^{-2}}{-2} = 14x^{-2}$
* Rewrite as fraction: $\frac{14}{x^2}$
* Result: $\frac{14}{x^2} + C$

7) $\int (\frac{12}{169x^{\frac{3}{13}}})dx$
* Rewrite: $\frac{12}{169} \int x^{-\frac{3}{13}} dx$
* Add 1 to exponent: $-\frac{3}{13} + \frac{13}{13} = \frac{10}{13}$
* Divide by new exponent: Multiply by the reciprocal ($\frac{13}{10}$).
* Calculation: $\frac{12}{169} \cdot \frac{13}{10} = \frac{12 \cdot 13}{169 \cdot 10}$. Since $169 = 13 \cdot 13$, we cancel one 13.
* Simplify: $\frac{12}{13 \cdot 10} = \frac{12}{130} = \frac{6}{65}$.
* Combine with variable: $\frac{6}{65}x^{\frac{10}{13}}$
* Result: $\frac{6}{65}x^{\frac{10}{13}} + C$

8) $\int (\frac{-3}{11x^{\frac{7}{11}}})dx$
* Rewrite: $-\frac{3}{11} \int x^{-\frac{7}{11}} dx$
* Add 1 to exponent: $-\frac{7}{11} + \frac{11}{11} = \frac{4}{11}$
* Divide by new exponent: Multiply by the reciprocal ($\frac{11}{4}$).
* Calculation: $-\frac{3}{11} \cdot \frac{11}{4}$. The 11s cancel out.
* Simplify: $-\frac{3}{4}$.
* Combine with variable: $-\frac{3}{4}x^{\frac{4}{11}}$
* Result: $-\frac{3}{4}x^{\frac{4}{11}} + C$

9) $\int (2x + 5)dx$
* Term 1: $\int 2x dx = x^2$
* Term 2: $\int 5 dx = 5x$
* Result: $x^2 + 5x + C$

10) $\int (\frac{-10}{133x^7})dx$
* Rewrite: $-\frac{10}{133} \int x^{-7} dx$
* Add 1 to exponent: $-7 + 1 = -6$
* Divide by new exponent: Divide by $-6$ (or multiply by $-\frac{1}{6}$).
* Calculation: $-\frac{10}{133} \cdot -\frac{1}{6} = \frac{10}{133 \cdot 6} = \frac{10}{798}$.
* Simplify fraction: Divide top and bottom by 2 $\rightarrow \frac{5}{399}$.
* Combine with variable: $\frac{5}{399}x^{-6}$ or $\frac{5}{399x^6}$.
* Result: $\frac{5}{399x^6} + C$

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Final Answer:
1) $x^2 - x + C$
2) $-x^4 + 12x^3 - 36x^2 + C$
3) $x^2 + 3x + C$
4) $\frac{1}{x} + C$
5) $\frac{19}{x^3} + C$
6) $\frac{14}{x^2} + C$
7) $\frac{6}{65}x^{\frac{10}{13}} + C$
8) $-\frac{3}{4}x^{\frac{4}{11}} + C$
9) $x^2 + 5x + C$
10) $\frac{5}{399x^6} + C$