You CAN Ace Calculus
related topics on other pages |
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For discussion of area under a parametric curve, see the parametric calculus page. |
For discussion of area under a polar curve, see the polar calculus page. |
external links you may find helpful |
Single Variable Calculus |
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Multi-Variable Calculus |
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Acceleration Vector |
Arc Length (Vector Functions) |
Arc Length Function |
Arc Length Parameter |
Conservative Vector Fields |
Cross Product |
Curl |
Curvature |
Cylindrical Coordinates |
Lagrange Multipliers |
Line Integrals |
Partial Derivatives |
Partial Integrals |
Path Integrals |
Potential Functions |
Principal Unit Normal Vector |
Differential Equations |
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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
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Help Keep 17Calculus Free |
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Area Under A Curve |
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To find the area under a curve that is above the x-axis, you just need to integrate the curve between two specific points. This is a natural idea from definite integrals. Here are a couple of videos explaining this.
video by rootmath
video by rootmath
Okay, to really understand this idea, let's work these practice problems.
Unless otherwise instructed, calculate the area under these curves, between the two points, if given.
\(y=x^2 ~~ [0,4]\)
Problem Statement |
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\(y=x^2 ~~ [0,4]\)
Solution |
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video by Michel vanBiezen
close solution |
\(y=x^3 ~~ [1,3]\)
Problem Statement |
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\(y=x^3 ~~ [1,3]\)
Solution |
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video by Michel vanBiezen
close solution |
\( y=1/x^2 ~~ [1,4] \)
Problem Statement |
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\( y=1/x^2 ~~ [1,4] \)
Solution |
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video by Michel vanBiezen
close solution |
\( y=\sin x ~~ [0,\pi] \)
Problem Statement |
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\( y=\sin x ~~ [0,\pi] \)
Solution |
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video by Michel vanBiezen
close solution |
\(y=x^2-2x+8 ~~ [1,2]\)
Problem Statement |
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\(y=x^2-2x+8 ~~ [1,2]\)
Solution |
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video by Michel vanBiezen
close solution |
\( y=\sin x \cos x \) \( [\pi/4, \pi/2] \)
Problem Statement |
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\( y=\sin x \cos x \) \( [\pi/4, \pi/2] \)
Solution |
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video by Michel vanBiezen
close solution |
\(\displaystyle{ y = \frac{x^2}{\sqrt{x^3+9}} ~~ [-1,1] }\)
Problem Statement |
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\(\displaystyle{ y = \frac{x^2}{\sqrt{x^3+9}} ~~ [-1,1] }\)
Solution |
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video by Michel vanBiezen
close solution |
When Area Under The Curve is Zero or Negative |
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Logically, area should always be positive. However, there will be times when you work the problem correctly and you end up with a negative number. What's the deal?
That's easy. When the area is below the x-axis, the area will turn out to be negative. So it is not that we have negative area. The sign tells us where MOST of the area is, above the x-axis for positive area or below the x-axis for negative area.
It is also possible to get zero for the area. This occurs when the area above the x-axis has exactly the same area as the area below the x-axis.
Here is a good video explaining in more detail why the area is negative. He uses an example and explains this very well.
From this discussion, you now understand why knowing what the graphs look like is so important. Let's work these practice problems. Some of these will have negative or zero area. Make sure you understand why.
Calculate the area under the curve \( y = 60x - 6x^2 \) on \( 0 \leq x \leq 15 \).
Problem Statement |
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Calculate the area under the curve \( y = 60x - 6x^2 \) on \( 0 \leq x \leq 15 \).
Final Answer |
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Problem Statement |
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Calculate the area under the curve \( y = 60x - 6x^2 \) on \( 0 \leq x \leq 15 \).
Solution |
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video by Michel vanBiezen
Final Answer |
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0 |
close solution |
Calculate the area under the curve \(y=x^2-6x\) between 1 and 3.
Problem Statement |
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Calculate the area under the curve \(y=x^2-6x\) between 1 and 3.
Final Answer |
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\(-46/3\) |
Problem Statement |
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Calculate the area under the curve \(y=x^2-6x\) between 1 and 3.
Solution |
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Final Answer |
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\(-46/3\) |
close solution |