Simplify the product.
2 answers:
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Integrals calculate the area underneath the curve. So, look at the graph, note where the function is on the graph and how many units on the X-axis the function is wide and how many units on the Y-axis the function is tall. The x-axis is usually where you would set the boundaries of the function. The integrand (the thing that we integrate) is the the function typically, and it’s in respect to something else, dx
, dy or dt to name a few. The thing we integrating with respect to, is dependent upon the graph.
Here’s an example: (Picture at the bottom)
If I take my bounds from 0 to 2 on the X-axis, that would give me an area underneath the curve that is exactly like a triangle. You can solve it with half base times height if you wanted.
If I were to take my bounds to be 2 to 6 on the X-axis then that’s an area in the shape of a trapezoid. Remember that you’re integrating the function so that means that your integrand will be the function.
So, for the first one it would be the integral sign, 2 on the top, 0 on the bottom, the integrand would be f(x), and it’s in respect to dt.
For the second one it would be exactly the same except your bounds would just be from 2 to 6
(2 on the bottom 6 on top).
I hope that helps.
, dy or dt to name a few. The thing we integrating with respect to, is dependent upon the graph.
Here’s an example: (Picture at the bottom)
If I take my bounds from 0 to 2 on the X-axis, that would give me an area underneath the curve that is exactly like a triangle. You can solve it with half base times height if you wanted.
If I were to take my bounds to be 2 to 6 on the X-axis then that’s an area in the shape of a trapezoid. Remember that you’re integrating the function so that means that your integrand will be the function.
So, for the first one it would be the integral sign, 2 on the top, 0 on the bottom, the integrand would be f(x), and it’s in respect to dt.
For the second one it would be exactly the same except your bounds would just be from 2 to 6
(2 on the bottom 6 on top).
I hope that helps.