C2 INTEGRATION Worksheet A - Free Printable

Here are the solutions to the problems on the worksheet. I have worked through each one step-by-step.

1. Integrate with respect to $x$

*Rule: Add 1 to the power, then divide by the new power.*

a) $\int x^5 \, dx = \frac{x^6}{6}$
b) $\int x^4 \, dx = \frac{x^5}{5}$
c) $\int x \, dx = \frac{x^2}{2}$
d) $\int x^4 \, dx = \frac{x^5}{5}$ *(Note: The image text for d is blurry but looks like $x^4$. If it is $x^{-4}$, answer is $\frac{x^{-3}}{-3}$)*. Let's assume standard positive integer based on context of 'a' and 'b'. Actually, looking closely at crop 1, d is likely $x^4$ or similar. Let's look at e: $5$. Integral of a constant $k$ is $kx$. So $\int 5 \, dx = 5x$.
f) $\int 3x^2 \, dx = 3(\frac{x^3}{3}) = x^3$
g) $\int 4x^7 \, dx = 4(\frac{x^8}{8}) = \frac{1}{2}x^8$
h) $\int 6x^{-2} \, dx = 6(\frac{x^{-1}}{-1}) = -6x^{-1}$ or $-\frac{6}{x}$
i) $\int 8x^3 \, dx = 8(\frac{x^4}{4}) = 2x^4$
j) $\int \frac{1}{4}x \, dx = \frac{1}{4}(\frac{x^2}{2}) = \frac{1}{8}x^2$
k) $\int 2x^{-9} \, dx = 2(\frac{x^{-8}}{-8}) = -\frac{1}{4}x^{-8}$ or $-\frac{1}{4x^8}$
l) $\int \frac{1}{5}x^{-3} \, dx = \frac{1}{5}(\frac{x^{-2}}{-2}) = -\frac{1}{10}x^{-2}$ or $-\frac{1}{10x^2}$

2. Find the integrals (Don't forget $+ C$)

a) $\int (2x + 3) \, dx = x^2 + 3x + C$
b) $\int (12x^2 - 4x) \, dx = \frac{12x^3}{3} - \frac{4x^2}{2} + C = 4x^3 - 2x^2 + C$
c) $\int (7 - x^3) \, dx = 7x - \frac{x^4}{4} + C$
d) $\int (x^2 + x + 1) \, dx = \frac{x^3}{3} + \frac{x^2}{2} + x + C$
e) $\int (x^4 + 5x^3) \, dx = \frac{x^5}{5} + \frac{5x^4}{4} + C$
f) $\int x(x^2 - 3) \, dx = \int (x^3 - 3x) \, dx = \frac{x^4}{4} - \frac{3x^2}{2} + C$
g) $\int (x - 2)^2 \, dx = \int (x^2 - 4x + 4) \, dx = \frac{x^3}{3} - 2x^2 + 4x + C$
h) $\int (3x^4 + x^2 - 6) \, dx = \frac{3x^5}{5} + \frac{x^3}{3} - 6x + C$
i) $\int (2 + \frac{1}{x^2}) \, dx = \int (2 + x^{-2}) \, dx = 2x + \frac{x^{-1}}{-1} + C = 2x - \frac{1}{x} + C$
j) $\int (x - \frac{1}{x^2}) \, dx = \int (x - x^{-2}) \, dx = \frac{x^2}{2} - \frac{x^{-1}}{-1} + C = \frac{x^2}{2} + \frac{1}{x} + C$
k) $\int x(\frac{2}{x^2} - 3) \, dx = \int (\frac{2}{x} - 3x) \, dx = 2\ln|x| - \frac{3x^2}{2} + C$
l) $\int (x - \frac{2}{x})^2 \, dx = \int (x^2 - 4 + \frac{4}{x^2}) \, dx = \frac{x^3}{3} - 4x - \frac{4}{x} + C$

3. Integrate with respect to $y$

a) $\int y^5 \, dy = \frac{y^6}{6}$
b) $\int y^{\frac{1}{2}} \, dy = \frac{y^{3/2}}{3/2} = \frac{2}{3}y^{3/2}$
c) $\int y^{-\frac{1}{2}} \, dy = \frac{y^{1/2}}{1/2} = 2y^{1/2}$ or $2\sqrt{y}$
d) $\int 4y^3 \, dy = y^4$
e) $\int y^{\frac{3}{2}} \, dy = \frac{y^{5/2}}{5/2} = \frac{2}{5}y^{5/2}$
f) $\int 5y^{-4} \, dy = 5(\frac{y^{-3}}{-3}) = -\frac{5}{3}y^{-3}$
g) $\int \sqrt{y} \, dy = \int y^{1/2} \, dy = \frac{2}{3}y^{3/2}$
h) $\int \frac{1}{\sqrt{y}} \, dy = \int y^{-1/2} \, dy = 2y^{1/2}$
i) $\int \frac{1}{2y^2} \, dy = \frac{1}{2}\int y^{-2} \, dy = \frac{1}{2}(-y^{-1}) = -\frac{1}{2y}$
j) $\int \sqrt{y^3} \, dy = \int y^{3/2} \, dy = \frac{2}{5}y^{5/2}$
k) $\int \frac{3}{y^2} \, dy = 3\int y^{-2} \, dy = 3(-y^{-1}) = -\frac{3}{y}$
l) $\int \frac{1}{\sqrt[3]{y}} \, dy = \int y^{-1/3} \, dy = \frac{y^{2/3}}{2/3} = \frac{3}{2}y^{2/3}$

4. Find the integrals

a) $\int (3t^2 - 1) \, dt = t^3 - t + C$
b) $\int (2r + \sqrt{r}) \, dr = \int (2r + r^{1/2}) \, dr = r^2 + \frac{2}{3}r^{3/2} + C$
c) $\int (3p - 1)^2 \, dp = \int (9p^2 - 6p + 1) \, dp = 3p^3 - 3p^2 + p + C$
d) $\int (4s + s^2) \, ds = 2s^2 + \frac{s^3}{3} + C$
e) $\int (\frac{1}{y^2} + y) \, dy = \int (y^{-2} + y) \, dy = -\frac{1}{y} + \frac{y^2}{2} + C$
f) $\int (\frac{1}{2}x^2 - x^3) \, dx = \frac{1}{2}(\frac{x^3}{3}) - \frac{x^4}{4} + C = \frac{x^3}{6} - \frac{x^4}{4} + C$
g) $\int \frac{t^2 - 2t}{t} \, dt = \int (t - 2) \, dt = \frac{t^2}{2} - 2t + C$
h) $\int (r^2 - r^{-3}) \, dr = \frac{r^3}{3} - \frac{r^{-2}}{-2} + C = \frac{r^3}{3} + \frac{1}{2r^2} + C$
i) $\int \frac{4p^2 - p^2}{2p} \, dp = \int \frac{3p^2}{2p} \, dp = \int \frac{3}{2}p \, dp = \frac{3}{4}p^2 + C$
j) $\int (4 - y^2) \, dy = 4y - \frac{y^3}{3} + C$
k) $\int \frac{1 + 6x^2}{3x^2} \, dx = \int (\frac{1}{3}x^{-2} + 2) \, dx = \frac{1}{3}(-x^{-1}) + 2x + C = -\frac{1}{3x} + 2x + C$
l) $\int \frac{2t + 3}{\sqrt{t}} \, dt = \int (2t^{1/2} + 3t^{-1/2}) \, dt = 2(\frac{2}{3}t^{3/2}) + 3(2t^{1/2}) + C = \frac{4}{3}t^{3/2} + 6t^{1/2} + C$

5. Find $\int y \, dx$

a) $y = 3x^2 - x + 6 \rightarrow \int y \, dx = x^3 - \frac{x^2}{2} + 6x + C$
b) $y = x^3 - x^2 + 2x - 5 \rightarrow \int y \, dx = \frac{x^4}{4} - \frac{x^3}{3} + x^2 - 5x + C$
c) $y = x(x - 2)(x + 1) = x(x^2 - x - 2) = x^3 - x^2 - 2x \rightarrow \int y \, dx = \frac{x^4}{4} - \frac{x^3}{3} - x^2 + C$
d) $y = (x^2 + 2)^2 = x^4 + 4x^2 + 4 \rightarrow \int y \, dx = \frac{x^5}{5} + \frac{4x^3}{3} + 4x + C$
e) $y = (x^2 - 4)(2x + 3) = 2x^3 + 3x^2 - 8x - 12 \rightarrow \int y \, dx = \frac{x^4}{2} + x^3 - 4x^2 - 12x + C$
f) $y = x^2 - 2x^2 + \frac{1}{x^2} = -x^2 + x^{-2} \rightarrow \int y \, dx = -\frac{x^3}{3} - \frac{1}{x} + C$
g) $y = \frac{1}{4x^2} - \frac{2}{x^3} = \frac{1}{4}x^{-2} - 2x^{-3} \rightarrow \int y \, dx = -\frac{1}{4}x^{-1} - 2(\frac{x^{-2}}{-2}) + C = -\frac{1}{4x} + \frac{1}{x^2} + C$
h) $y = (1 - \frac{2}{x})^2 = 1 - \frac{4}{x} + \frac{4}{x^2} \rightarrow \int y \, dx = x - 4\ln|x| - \frac{4}{x} + C$
i) $y = (x^2 - 1)(x^2 + 1) = x^4 - 1 \rightarrow \int y \, dx = \frac{x^5}{5} - x + C$

6. Find a general expression for $y$ given $\frac{dy}{dx}$

*(This means integrate the function and add $+ C$)*

a) $\frac{dy}{dx} = 8x + 3 \rightarrow y = 4x^2 + 3x + C$
b) $\frac{dy}{dx} = \frac{1}{2}x^2 - x^3 \rightarrow y = \frac{1}{6}x^3 - \frac{1}{4}x^4 + C$
c) $\frac{dy}{dx} = \frac{4}{x^2} = 4x^{-2} \rightarrow y = -4x^{-1} + C = -\frac{4}{x} + C$
d) $\frac{dy}{dx} = (x + 1)^2 = x^2 + 2x + 1 \rightarrow y = \frac{x^3}{3} + x^2 + x + C$
e) $\frac{dy}{dx} = 2x - \frac{1}{x^2} = 2x - x^{-2} \rightarrow y = x^2 - (-x^{-1}) + C = x^2 + \frac{1}{x} + C$
f) $\frac{dy}{dx} = x^2 - 2x^{-4} \rightarrow y = \frac{x^3}{3} - 2(\frac{x^{-3}}{-3}) + C = \frac{x^3}{3} + \frac{2}{3x^3} + C$
g) $\frac{dy}{dx} = \frac{3 - x^2}{2x^4} = \frac{3}{2}x^{-4} - \frac{1}{2}x^{-2} \rightarrow y = \frac{3}{2}(\frac{x^{-3}}{-3}) - \frac{1}{2}(\frac{x^{-1}}{-1}) + C = -\frac{1}{2x^3} + \frac{1}{2x} + C$
h) $\frac{dy}{dx} = \frac{2}{x}(5 - x) = \frac{10}{x} - 2 \rightarrow y = 10\ln|x| - 2x + C$
i) $\frac{dy}{dx} = \frac{9x - 2}{\sqrt{x}} = 9x^{1/2} - 2x^{-1/2} \rightarrow y = 9(\frac{2}{3}x^{3/2}) - 2(2x^{1/2}) + C = 6x^{3/2} - 4x^{1/2} + C$

Final Answer:
See the detailed solutions above for questions 1 through 6.