Everything from the 3 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
\(\displaystyle\int x^{n}\,dx = \dfrac{x^{n+1}}{n+1} + c\) and \(\displaystyle\int (ax + b)^{n}\,dx = \dfrac{(ax + b)^{n+1}}{a(n+1)} + c\), for any rational \(n \ne -1\). A point on the curve fixes the value of \(c\).
Dividing by a fraction is multiplying by its reciprocal: \(\dfrac{x^{\frac{3}{2}}}{\frac{3}{2}} = \tfrac{2}{3}x^{\frac{3}{2}}\), not \(\tfrac{3}{2}x^{\frac{3}{2}}\). And \(\displaystyle\int x^{-1}\,dx\) cannot be done this way (it would divide by zero).
Divide by \(a\); do not multiply. \(\displaystyle\int (2x + 1)^{3}\,dx\) is \(\dfrac{(2x + 1)^{4}}{8}\), not \(\dfrac{2(2x + 1)^{4}}{4}\). This shortcut only works when the inside is LINEAR: \(\displaystyle\int (x^{2} + 1)^{3}\,dx\) needs expanding.
Find \(c\) straight after integrating, using the equation of the CURVE (\(y\)), not the gradient. When you integrate twice, each integration needs its own constant.
\(\displaystyle\int_{a}^{b} f(x)\,dx = F(b) - F(a)\), where \(F\) is any integral of \(f\). The area between \(y = f(x)\) and the \(x\)-axis is \(\displaystyle\int_{a}^{b} y\,dx\) when the curve is above the axis; between two graphs, integrate (top \(-\) bottom).
A lower limit of 0 does not always give 0: \((2 \times 0 + 1)^{\frac{3}{2}} = 1\). Always substitute properly, and put the bottom-limit value in brackets before subtracting it.
Sketch first. If the curve crosses the \(x\)-axis between the limits, one integral gives (area above) \(-\) (area below), not the total area.
Upper minus lower. If you subtract the wrong way you get the negative of the area; an area is never negative, so take the size, but check your sketch.
An improper integral only has a value if the integrated expression settles to a limit. \(\left[-\tfrac{1}{x}\right]\) at \(\infty\) is 0, but something like \(\left[2\sqrt{x}\right]\) at \(\infty\) grows without limit, so \(\displaystyle\int_{1}^{\infty} \dfrac{1}{\sqrt{x}}\,dx\) has no value.
About the \(x\)-axis: \(V = \pi\displaystyle\int_{a}^{b} y^{2}\,dx\). About the \(y\)-axis: \(V = \pi\displaystyle\int_{c}^{d} x^{2}\,dy\). For a region with a hole, subtract: \(V = \pi\displaystyle\int_{a}^{b} \left(y_{\text{outer}}^{2} - y_{\text{inner}}^{2}\right) dx\).
Square \(y\) BEFORE integrating: \(\pi\displaystyle\int y^{2}\,dx\) is not \(\pi\left(\displaystyle\int y\,dx\right)^{2}\). And keep the \(\pi\): an exact answer is usually wanted, such as \(\tfrac{206\pi}{15}\).
About the \(y\)-axis, the limits are \(y\)-values. Using \(x = 1\) to \(x = 2\) (where the curve meets \(y = 1\) and \(y = 4\)) with \(\displaystyle\int x^{2}\,dy\) mixes the two variables.
Washers: subtract the SQUARES, \(\pi\displaystyle\int (y_{1}^{2} - y_{2}^{2})\,dx\). Squaring the difference, \(\pi\displaystyle\int (y_{1} - y_{2})^{2}\,dx\), rotates the wrong shape and is the most common error.