AP Calculus AB/BC

Dec 14, 2025

152

4 pages

Understanding Riemann Sums and Definite Integrals

Chelsea @zuncho

Riemann sums and definite integrals are powerful tools that help us find the area under curves. Whether you're... Show more

5
Section 4.4: Riemann sum 3 Definite Integrals.

Note: Riemann sums help estimate the area under curve.

Notes Types of Riemann sums:

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Riemann Sums Estimating Areas

Ever wonder how to measure the area under a curved line? Riemann sums break the problem into manageable pieces! These sums estimate area by dividing regions into rectangles or trapezoids.

There are four main types of Riemann sums you need to know Left, Right, Midpoint, and Trapezoidal. Each method places rectangles at different positions to approximate the area under a curve.

Let's see how a Left Riemann Sum works with an example For $f(x)=x^2+1$ from $[0,8]$ with 4 sub-intervals, we first calculate the width of each interval $\Delta x = \frac{8-0}{4} = 2$ . Then we calculate the sum of all rectangle areas $2 \cdot f(0) + 2 \cdot f(2) + 2 \cdot f(4) + 2 \cdot f(6) = 120$ .

Pro Tip On the AP Calculus test, you can leave your Riemann sum fully written out without calculating the final answer—this is acceptable and saves time!

More Riemann Sum Examples

The placement of your rectangles makes a big difference in your estimate's accuracy. With a Midpoint Riemann Sum, we evaluate the function at the middle of each sub-interval.

For example, approximating the area under $f(x) = x^2 + 1$ from $x = -3$ to $x = 3$ with 3 equal sub-intervals gives us $2 \cdot f(-2) + 2 \cdot f(0) + 2 \cdot f(2) = 2(5) + 2(1) + 2(5) = 22$ .

The Trapezoidal Riemann sum often gives better approximations because it uses trapezoids instead of rectangles. The area of a trapezoid is calculated as $\frac{1}{2}(b_1 + b_2) \cdot h$ , where $b_1$ and $b_2$ are the bases and $h$ is the height.

When working with Riemann sums, remember that Left and Right sums can over- or underestimate the true area. Midpoint and Trapezoidal sums typically give more accurate approximations because they account for the curve's behavior better.

Definite Integrals Finding Exact Areas

While Riemann sums give us approximations, definite integrals calculate the exact area under a curve. We write this as $\int_{a}^{b} f(x) dx$ , where $a$ and $b$ are the endpoints of our interval.

One way to find definite integrals is by using familiar area formulas for common shapes. You can break complex regions into basic geometric shapes like rectangles, triangles, trapezoids, circles, and semicircles.

For example, to find $\int_{1}^{3} 4 dx$ , we recognize this as the area of a rectangle with height 4 and width 2, giving us 8. For $\int_{0}^{3} (x+2) dx$ , we can identify a trapezoid shape with bases 2 and 5, and height 3.

Remember When calculating areas of curves like $\int_{-2}^{2} \sqrt{4-x^2} dx$ , recognize that this represents a semicircle with radius 2, making the answer $\frac{\pi \cdot 2^2}{2} = 2\pi$ .

Properties of Definite Integrals

Definite integrals have important properties that make calculations easier. If you integrate over an interval with the same start and endpoint $$a=b$$ , the result is always zero $\int_{a}^{a} f(x) dx = 0$ .

When you reverse the limits of integration, you simply negate the result $\int_{b}^{a} f(x) dx = -\int_{a}^{b} f(x) dx$ . This makes sense if you think about traveling backward along the same path.

The Additive Interval Property is super helpful $\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx$ where $a \leq c \leq b$ . This lets you break a difficult integral into more manageable pieces.

For example, to find $\int_{-1}^{1} |x| dx$ , we can split it at x=0 where the function changes behavior $\int_{-1}^{0} |x| dx + \int_{0}^{1} |x| dx = \frac{1}{2} + \frac{1}{2} = 1$ . This property also helps with problem-solving when you know the values of certain integrals already.

Quick Check If you know that $\int_{0}^{7} f(x) dx = 8$ and $\int_{1}^{7} f(x) dx = -4$ , you can find $\int_{0}^{1} f(x) dx = 12$ using the additive property.

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AP Calculus AB/BC

•

152

•

Dec 14, 2025

•

4 pages

Understanding Riemann Sums and Definite Integrals

Chelsea

@zuncho

Riemann sums and definite integrals are powerful tools that help us find the area under curves. Whether you're calculating the area beneath a roller coaster path or understanding calculus basics, these concepts are essential for solving real-world problems involving changing... Show more

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Riemann Sums: Estimating Areas

Ever wonder how to measure the area under a curved line? Riemann sums break the problem into manageable pieces! These sums estimate area by dividing regions into rectangles or trapezoids.

There are four main types of Riemann sums you need to know: Left, Right, Midpoint, and Trapezoidal. Each method places rectangles at different positions to approximate the area under a curve.

Let's see how a Left Riemann Sum works with an example: For $f(x)=x^2+1$ from $[0,8]$ with 4 sub-intervals, we first calculate the width of each interval: $\Delta x = \frac{8-0}{4} = 2$ . Then we calculate the sum of all rectangle areas: $2 \cdot f(0) + 2 \cdot f(2) + 2 \cdot f(4) + 2 \cdot f(6) = 120$ .

Pro Tip: On the AP Calculus test, you can leave your Riemann sum fully written out without calculating the final answer—this is acceptable and saves time!

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More Riemann Sum Examples

The placement of your rectangles makes a big difference in your estimate's accuracy. With a Midpoint Riemann Sum, we evaluate the function at the middle of each sub-interval.

For example, approximating the area under $f(x) = x^2 + 1$ from $x = -3$ to $x = 3$ with 3 equal sub-intervals gives us: $2 \cdot f(-2) + 2 \cdot f(0) + 2 \cdot f(2) = 2(5) + 2(1) + 2(5) = 22$ .

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Definite Integrals: Finding Exact Areas

Remember: When calculating areas of curves like $\int_{-2}^{2} \sqrt{4-x^2} dx$ , recognize that this represents a semicircle with radius 2, making the answer $\frac{\pi \cdot 2^2}{2} = 2\pi$ .

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Properties of Definite Integrals

When you reverse the limits of integration, you simply negate the result: $\int_{b}^{a} f(x) dx = -\int_{a}^{b} f(x) dx$ . This makes sense if you think about traveling backward along the same path.

The Additive Interval Property is super helpful: $\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx$ where $a \leq c \leq b$ . This lets you break a difficult integral into more manageable pieces.

For example, to find $\int_{-1}^{1} |x| dx$ , we can split it at x=0 where the function changes behavior: $\int_{-1}^{0} |x| dx + \int_{0}^{1} |x| dx = \frac{1}{2} + \frac{1}{2} = 1$ . This property also helps with problem-solving when you know the values of certain integrals already.

Quick Check: If you know that $\int_{0}^{7} f(x) dx = 8$ and $\int_{1}^{7} f(x) dx = -4$ , you can find $\int_{0}^{1} f(x) dx = 12$ using the additive property.