Integration Revision 1
Finding the area under a curve given the equation between two points Ben smith
Start by integrating the function
• And put in square brackets
Now replace the x values with the numbers on the edge of the square brackets
• =(32+512)-(8+64) • =472
If an area between the x-axis and the curve is both above and below the x-axis • The problem we face when doing this is
that if we simply input values above and below the x-axis we get a positive value and a negative value. We don t get the correct area as the negative needs to be added in it s positive form in order to find the true area
Example • Y=X3 • Find the area between -2 and 2 • For this if we simply put it in the method
we will get an answer of 0 • Instead we have to find the area for 0 and 2 and 0 and -2 • Both give us the answer 4 so the final answer
Quick point
• We can find where the graph is
below and above the axis and therefore what points to take by simply finding the roots of the equation
Start by integrating the function
• And put in square brackets
Now replace the x values with the numbers on the edge of the square brackets
• =(32+512)-(8+64) • =472
If an area between the x-axis and the curve is both above and below the x-axis • The problem we face when doing this is
that if we simply input values above and below the x-axis we get a positive value and a negative value. We don t get the correct area as the negative needs to be added in it s positive form in order to find the true area
Example • Y=X3 • Find the area between -2 and 2 • For this if we simply put it in the method
we will get an answer of 0 • Instead we have to find the area for 0 and 2 and 0 and -2 • Both give us the answer 4 so the final answer
Quick point
• We can find where the graph is
below and above the axis and therefore what points to take by simply finding the roots of the equation
