Area L

Draw the graph of a continuous increasing function in the first quadrant and horizontal and vertical lines through two points. The areas in your sketch lead to a useful formula for finding integrals.

Integral Equation

Solve this integral equation.

Integral Sandwich

Generalise this inequality involving integrals.

Calculus Countdown

Stage: 5 Challenge Level:

Game a): $\textrm{Target} = 8$

$\mathrm{D} \left( x^2 \right) = 2x$

$\mathrm{D} \left( 2x \right) = 2$

$\mathrm{P} \left( 4,2 \right) = 8$

Game b): $\textrm{Target} = x^4$

$\mathrm{P} \left( x, x^2 \right) = x^3$

$\mathrm{I} \left( x^3 \right) = \frac{x^4}{4}$

$\mathrm{P} \left( \frac{x^4}{4}, 4 \right) = x^4$

Game c): $\textrm{Target} = \frac{1}{2}$

$\mathrm{D} \left( x^2 \right) = 2x$

$\mathrm{D} \left( 2x \right) = 2$

$\mathrm{R} \left( 2 \right) = \frac{1}{2}$

Game d): $\textrm{Target} = \frac{x^6}{36}$

Method 1:

$\mathrm{I} \left( \mathrm{I}(x) \right) = \frac{x^3}{6}$

$\mathrm{P} \left( x^2,\frac{x^3}{6} \right) = \frac{x^5}{6}$

$\mathrm{I} \left( \frac{x^5}{6} \right) = \frac{x^6}{36}$

Method 2:

$\mathrm{I} \left( x^2 \right) = \frac{x^3}{3}$

$\mathrm{D} \left( \ln(x) \right) = \frac{1}{x}$

$\mathrm{R} \left( \frac{1}{x} \right) = x$

$\mathrm{P} \left( x, x \right) = x^2$

$\mathrm{I} \left( x^2 \right) = \frac{x^3}{3}$

$\mathrm{P} \left( \frac{x^3}{3}, \frac{x^3}{3} \right) = \frac{x^6}{9}$

$\mathrm{R} \left( 4 \right) = 0.25$

$\mathrm{P} \left( 0.25,\frac{ x^6}{9} \right) =\frac{x^6}{36}$

Game e): $\textrm{Target} = \frac{-32}{x^5}$

$\mathrm{D} \left( \mathrm{D} \left( \mathrm{D} \left( \ln(x) \right) \right) \right) = 2x^{-3}$

$\mathrm{R} \left( x \right) = \frac{1}{x}$

$\mathrm{P} \left( x^{-1}, 2x^{-3} \right) = 2x^{-4}$

$\mathrm{P} \left( 4, 2x^{-4} \right) = 8x^{-4}$

$\mathrm{D} \left( 8x^{-4} \right) = -32x^{-5}$

Game f): $\textrm{Target} = x(2 - x)$

$\mathrm{P} \left( x^2, \mathrm{R}(\exp(x)) \right) =x^2 \exp(-x))$

$\mathrm{D} \left( x^2 \exp(-x) \right) = 2x \exp(-x) - x^2 \exp(-x)$

$\mathrm{P} \left( \exp(x),2x \exp(-x) - x^2 \exp(-x) \right) = 2x-x^2$