AREA |

**Area Under a Curve**

In the lesson on Riemann sums, we learned that the area between a curve and the *x*-axis is found by evaluating where *f *(*x _{i}*) is the height and is the base of the

If the curve is always above the *x*-axis, we simply integrate from *a* to *b*. However, **if the curve** of the function **crosses the x-axis**, we must

**Example 1:** Find the area between *f* (*x*)* = x*² + 3 and the *x*-axis from *x* = 0 to *x* = 4.

Since the curve doesn't cross the *x*-axis, we simply evaluate the integral from *x* = 0 to *x* = 4.

So, the area is

Since the value = 0 at *x* = 0, we substitute *x* = 4

The area under this curve from *x* = 0 to *x* = 4 is 48 units².

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**Example 2:** In this one, we have to break the area up into 2 sections since the curve crosses the *x*-axis on the **interval of integration**.

Find the area between the *x*-axis and the curve* **f *(*x*)* = x*² – 4 from *x* = 0 to *x* = 4.

Here's the picture

Because this curve crosses the *x*-axis at *x* = 2, making the ** y-values negative**, we must integrate the

In this case,

This is really

Once we integrate (antidifferentiate), we get

again, at *x* = 0 everything disappears so,

The area between this curve and the *x*-axis from *x* = 0 to *x* = 4 is 16 units².

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**Area Between Two Curves**

When we have to find the area **between two curves**, we must pay attention to **which curve forms** the top or **ceiling** and **which forms** the bottom or **floor** of the region between them. We must also **find their points of intersection** to determine the limits of integration.

**Example 3:** Evaluate the area between *y* = 9 – 2*x*² and *y* = – *x*².

Setting *y* = *y*, we learn that the points of intersection are at *x* = ± 3.

Here's the picture

Since the curves are symmetric to the *y*-axis, we need only **find the area on one side** and then multiply it **by 2**. We'll take **twice the area on the right side** of the axis so we'll integrate from *x* = 0 to *x* = 3.

The blue curve, *y* = 9 – 2*x*² is the ceiling and

the red curve *y* = – *x*² is the floor, so our area is:

Once we antidifferentiate and evaluate, we get that the Area = 48 units².

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**Area of a Region Defined by More Than Two Curves**

Now let's do an example where we're given 3 different functions and we want to find the area of the region enclosed between them.

**Example 4:** Find the area of the region defined by *y* = *x*, *y* = 3*x*, and *y* = – *x* + 4.

Here's the picture:

The line *y* = *x* is always the floor.

At first, the ceiling is *y *= 3*x* from *x* = 0 to *x* = 1

right of *x* = 1, *y* = – *x* + 4** **is the ceiling up to *x* = 2.

Therefore, the area here is found by evaluating

the area is 2 square units.

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Sometimes, the functions are such that it is **impossible to use the x-axis as reference** to complete the question because the

**Example 5:** Find the area of the region bounded by *x* = *y*² and *x* = – *y*² + 8.

Here's the picture:

When we set *x* = *x*, we get that *y* = ± 2, and since everything is **symmetric to the x-axis**, we will evaluate

The "ceiling" is the blue curve, the "floor" is the red one.

This becomes

the area is 64/3 square units.

Now get a pencil, an eraser and a note book, copy the questions,

do the practice exercise(s), then check your work with the solutions.

If you get stuck, review the examples in the lesson, then try again.

**Practice**

Sketch the region bounded by the graphs of these functions, then find the area of the region.

1) y = 1/x² |
y = – x² |
from x = 1 to x = 2 |

2) | y = – x |
from x = 1 to x = 4 |

3) y² = – x |
x – y = 4 |
from y = – 1 to y = 2 |

4) y² = x |
y – x = 2 |
from y = – 2 to y = 3 |

5) x – y + 1 = 0 |
7 |
2 |

**Solutions**

1) *y* = 1/*x*², ................ *y* = – *x*², .......... *x* = 1, .......... and .......... *x* = 2

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2) , .............. *y* = – *x*, ........... *x* = 1, ............ and .......... *x* = 4

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3) *y*² = – *x* , ................*x* – *y* = 4 ......... *y* = – 1, .......... and .......... *y* = 2

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4) *y*² = *x* , ..................*y* – *x* = 2 ......... *y* = – 2, .......... and .......... *y* = 3

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5) *x* – *y* + 1 = 0 ........ 7*x* – *y* – 17 = 0 .......... 2*x* + *y* + 2 = 0

When we put these equations in standard form we get:

*y* = *x *+ 1........ *y* = 7*x* – 17.......... and *y* = – 2*x* – 2

( *Calculus II MathRoom Index* )

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