## 17Calculus Integrals - Area Under Curves

Area Under A Curve

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.

### rootmath - Intro to Area Under a Curve [4min-28secs]

video by rootmath

### rootmath - Exact Area Under A Curve [9min-55secs]

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

$$y=x^2 ~~ [0,4]$$

Solution

### 2114 video

video by Michel vanBiezen

$$y=x^3 ~~ [1,3]$$

Problem Statement

$$y=x^3 ~~ [1,3]$$

Solution

### 2115 video

video by Michel vanBiezen

$$y=1/x^2 ~~ [1,4]$$

Problem Statement

$$y=1/x^2 ~~ [1,4]$$

Solution

### 2117 video

video by Michel vanBiezen

$$y=\sin x ~~ [0,\pi]$$

Problem Statement

$$y=\sin x ~~ [0,\pi]$$

Solution

### 2118 video

video by Michel vanBiezen

$$y=x^2-2x+8 ~~ [1,2]$$

Problem Statement

$$y=x^2-2x+8 ~~ [1,2]$$

Solution

### 2116 video

video by Michel vanBiezen

$$y=\sin x \cos x$$ $$[\pi/4, \pi/2]$$

Problem Statement

$$y=\sin x \cos x$$ $$[\pi/4, \pi/2]$$

Solution

### 2119 video

video by Michel vanBiezen

$$\displaystyle{ y = \frac{x^2}{\sqrt{x^3+9}} ~~ [-1,1] }$$

Problem Statement

$$\displaystyle{ y = \frac{x^2}{\sqrt{x^3+9}} ~~ [-1,1] }$$

Solution

### 2120 video

video by Michel vanBiezen

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 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.

### Section 5.3 - Integrals and Negative Area [5min-21secs]

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

Calculate the area under the curve $$y = 60x - 6x^2$$ on $$0 \leq x \leq 15$$.

0

Problem Statement

Calculate the area under the curve $$y = 60x - 6x^2$$ on $$0 \leq x \leq 15$$.

Solution

### 2433 video

video by Michel vanBiezen

0

Calculate the area under the curve $$y=x^2-6x$$ between 1 and 3.

Problem Statement

Calculate the area under the curve $$y=x^2-6x$$ between 1 and 3.

$$-46/3$$

Problem Statement

Calculate the area under the curve $$y=x^2-6x$$ between 1 and 3.

Solution

### 2434 video

$$-46/3$$

You CAN Ace Calculus

 integration definite integrals

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.

### Calculus Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

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