Find $$\displaystyle\int x^4 + x^3\ \mathrm{d}x$$
$\dfrac{1}{5}x^5 + \dfrac{1}{4}x^4 + c$
Find $$\displaystyle\int 2x - x^2 + 5\ \mathrm{d}x$$
$x^2 - \dfrac{1}{3}x^3 + 5x + c$
Find $$\displaystyle\int x\sqrt{x}\ \mathrm{d}x$$
$\dfrac{2}{5}x^{\frac{5}{2}} + c$
Find $$\displaystyle\int \dfrac{1}{x^6}\ \mathrm{d}x$$
$-\dfrac{1}{5}x^{-5} + c$
Find $$\displaystyle\int \dfrac{1}{2x^2}\ \mathrm{d}x$$
$-\dfrac{1}{2}x^{-1} + c$
Find $$\displaystyle\int \dfrac{3}{2x^3}\ \mathrm{d}x$$
$-\dfrac{3}{4}x^{-2} + c$
Find $$\displaystyle\int \dfrac{x^3 + 1}{x^2}\ \mathrm{d}x$$
$\dfrac{1}{2}x^2 - \dfrac{1}{x} + c$
Find $$\displaystyle\int (x+2)^2\ \mathrm{d}x$$
$\dfrac{1}{3}x^3 + 2x^2 + 4x + c$
Find $$\displaystyle\int \dfrac{(2+x)(3x-1)}{\sqrt{x}}\ \mathrm{d}x$$
$\dfrac{6}{5}x^{\frac{5}{2}} + \dfrac{10}{3}x^{\frac{3}{2}} + 4x^{\frac{1}{2}} +c$
Find $$\displaystyle\int \sqrt{x}(2x+3\sqrt{x})\ \mathrm{d}x$$
$\dfrac{4}{5}x^{\frac{5}{2}} + \dfrac{3}{2}x^2 + c$
Find $y$ given $$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 + 2x$$ and $y = 10$ when $x = 2$
$y = x^3 + x^2 - 2$
Find $y$ given $$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x^3 + \dfrac{2}{x^2} + 3$$ and $y = 2$ when $x = 1$
$y = x^4 - \dfrac{2}{x} + 3x$
Find $y$ given $$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\sqrt{x} + \dfrac{1}{2}x^2$$ and $y = 64$ when $x = 4$
$y = \dfrac{4}{3}x^{\frac{3}{2}} + \dfrac{1}{6}x^3 + \dfrac{128}{3}$
Find $y$ given $$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sqrt{x}} - 3x$$ and $y = 4$ when $x = 0$
$y = 2\sqrt{x} - \dfrac{3}{2}x^2 + 4$
A curve has gradient given by $3 + 2x - x^2$ and passes through the point $(3, 5)$. Find the equation of the curve.
$y = 3x + x^2 - \dfrac{1}{3}x^3 - 4$
A curve has gradient given by $2x^3 - x - 8$ and passes through the point $(-1, 4)$. Find the equation of the tangent to the curve at the point where $x = 2$.
$y = 6x - 26$
A curve passes through the origin. Given $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 8x - 5$, find the coordinates of the other points where the curve crosses the $x$ axis.
$(-1, 0)$ and $(5, 0)$
A curve has gradient given by $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 2x - 3$, and passes through $(-2, 0)$. Find the equation of the tangent to the curve when $x = 1$.
$y = 5 - 2x$
The gradient of a curve is given by $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 + kx$. Given that the curve passes through $(1, 6)$ and $(2,1)$, find the equation of the curve.
$y = x^3 - 4x^2 + 9$
The gradient of a curve is given by $\dfrac{\mathrm{d}y}{\mathrm{d}x} = k - \dfrac{1}{\sqrt{x}}$. Given that the curve passes through $(1, -2)$ and $(4, 5)$, find the equation of the normal to the curve when $x = 1$. Give your answer in the form $ax + by + c = 0$.
$x + 2y + 3 = 0$