Area Under A Curve
To find the area under a curve that is above the xaxis, 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.
Practice
Unless otherwise instructed, calculate the area under these curves, between the two points, if given.
\( y=x^2 \); \( [0,4] \)
Problem Statement 

Calculate the area under the curve \( y=x^2 \) on the interval \( [0,4] \).
Solution 

video by Michel vanBiezen 

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\( y=x^3 \); \( [1,3] \)
Problem Statement 

Calculate the area under the curve \( y=x^3 \) on the interval \( [1,3] \).
Solution 

video by Michel vanBiezen 

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\( y=1/x^2 \); \( [1,4] \)
Problem Statement 

Calculate the area under the curve \( y=1/x^2 \) on the interval \( [1,4] \).
Solution 

video by Michel vanBiezen 

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\( y=\sin x \); \( [0,\pi] \)
Problem Statement 

Calculate the area under the curve \( y=\sin x \) on the interval \( [0,\pi] \).
Solution 

video by Michel vanBiezen 

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\(y=x^22x+8\); \([1,2]\)
Problem Statement 

Calculate the area under the curve \(y=x^22x+8\) on the interval \([1,2]\).
Solution 

video by Michel vanBiezen 

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\( y=\sin x \cos x \); \( [\pi/4, \pi/2] \)
Problem Statement 

Calculate the area under the curve \( y=\sin x \cos x \) on the interval \( [\pi/4, \pi/2] \).
Solution 

video by Michel vanBiezen 

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\(\displaystyle{ y = \frac{x^2}{\sqrt{x^3+9}} }\); \( [1,1] \)
Problem Statement 

Calculate the area under the curve \(\displaystyle{ y = \frac{x^2}{\sqrt{x^3+9}} }\) on the interval \( [1,1] \).
Solution 

video by Michel vanBiezen 

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When Area Under The Curve is Zero or Negative
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 xaxis, 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 xaxis for positive area or below the xaxis for negative area.
It is also possible to get zero for the area. This occurs when the area above the xaxis has exactly the same area as the area below the xaxis.
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.
Practice
Unless otherwise instructed, calculate the area under these curves, between the two points, if given. Some of these will have negative or zero area. Make sure you understand why.
\( y = 60x  6x^2 \); \( 0 \leq x \leq 15 \)
Problem Statement 

Calculate the area under the curve \( y = 60x  6x^2 \) on the interval \( 0 \leq x \leq 15 \). The area may be negative or zero. Make sure you understand why.
Solution 

video by Michel vanBiezen 

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\(y=x^26x\); \( [1,3] \)
Problem Statement 

Calculate the area under the curve \(y=x^26x\) on the interval \( [1,3] \). The area may be negative or zero. Make sure you understand why.
Final Answer 

\(46/3\)
Problem Statement 

Calculate the area under the curve \(y=x^26x\) on the interval \( [1,3] \). The area may be negative or zero. Make sure you understand why.
Solution 

Final Answer 

\(46/3\) 
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You CAN Ace Calculus
related topics on other pages 

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 
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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Practice Instructions
Unless otherwise instructed, calculate the area under these curves, between the two points, if given.