1.08b Integrate x^n: where n != -1 and sums

Find \(\int \left(x - \frac{3}{x^2}\right) dx\). [3]

Find \(\int (20x^4 + 6x^{-\frac{2}{3}}) dx\). [4]

Find \(\int (12x^5 + \sqrt[5]{x} + 7) dx\). [5]

Find \(\int \left(x^{\frac{1}{2}} + \frac{6}{x^3}\right) dx\). [5]

Find \(\int \left(x^4 + \frac{1}{x^3}\right) dx\). [4]

The function f, defined for \(x \in \mathbb{R}, x > 0\), is such that $$f'(x) = x^2 - 2 + \frac{1}{x^2}$$

1. Find the value of f''(x) at \(x = 4\). [3]
2. Given that f(3) = 0, find f(x). [4]
3. Prove that f is an increasing function. [3]

A cyclist travels along a straight road. Her velocity \(v\) m s\(^{-1}\), at time \(t\) seconds after starting from a point \(O\), is given by \(v = 2\) for \(0 \leq t \leq 10\), \(v = 0.03t^2 - 0.3t + 2\) for \(t \geq 10\).

1. Find the displacement of the cyclist from \(O\) when \(t = 10\). [1]
2. Show that, for \(t \geq 10\), the displacement of the cyclist from \(O\) is given by the expression \(0.01t^3 - 0.15t^2 + 2t + 5\). [4]
3. Find the time when the acceleration of the cyclist is \(0.6\) m s\(^{-2}\). Hence find the displacement of the cyclist from \(O\) when her acceleration is \(0.6\) m s\(^{-2}\). [5]

It is given that $$\frac{\mathrm{d}y}{\mathrm{d}x} = (x + 2)(2x - 1)^2$$ and when \(x = 6\), \(y = 900\) Find \(y\) in terms of \(x\) [6 marks]

Given that \(\frac{dy}{dx} = \frac{1}{6x^2}\), find \(y\). Circle your answer. \(\frac{-1}{3x^3} + c\) \quad \(\frac{1}{2x^3} + c\) \quad \(\frac{-1}{6x} + c\) \quad \(\frac{-1}{3x} + c\) [1 mark]

\(f'(x) = \left(2x - \frac{3}{x}\right)^2\) and \(f(3) = 2\) Find \(f(x)\). [4 marks]

Find $$\int \left(x^2 + x^{\frac{1}{2}}\right) dx$$ [2 marks]