Definite Integrals and Riemann Sums

Area Under a Curve and Riemann Sums

To find the area under a curve between points a and b, we can approximate it using rectangles.
Riemann sums involve constructing these rectangles and taking the limit as the width of the rectangles approaches zero.
As the rectangles become narrower, the approximation gets closer to the actual area under the curve.

The definite integral can be interpreted as the "signed area" under a curve.
Area above the x-axis is considered positive, while area below the x-axis is considered negative.
Given a function f(t), the definite integral from a to b is expressed as \int_{a}^{b} f(t) dt.
If part of the curve is below the x-axis (area B) and part is above (area A), the definite integral is A - B.

Consider finding the integral \int_{0}^{2} (1 - 2x) dx.
The function 1 - 2x is a line with a y-intercept of 1 and a slope of -2.
The area above the x-axis is a triangle with base \frac{1}{2} and height 1, yielding an area of \frac{1}{4}. Area is \frac{1}{2} \cdot base \cdot height = \frac{1}{2} \cdot \frac{1}{2} \cdot 1 = \frac{1}{4}.
The area below the x-axis is a triangle with base \frac{3}{2} and height 3, yielding an area of \frac{9}{4}. Area is \frac{1}{2} \cdot base \cdot height = \frac{1}{2} \cdot \frac{3}{2} \cdot 3 = \frac{9}{4}.
\The value of the definite integral is the area above minus the area below: \frac{1}{4} - \frac{9}{4} = -\frac{8}{4} = -2.

Problem: Find the integral \int_{1}^{3} (2 + 3x) dx.
The function 2 + 3x is a line.
The region of interest forms a trapezoid.
The area of a trapezoid is given by the average of the bases times the height: \frac{1}{2}(b1 + b2)h.
At x = 1, the function value is 2 + 3(1) = 5.
At x = 3, the function value is 2 + 3(3) = 11.
The height of the trapezoid is 3 - 1 = 2.
The area is \frac{1}{2}(5 + 11) \cdot 2 = 16.

The integral \int{1}^{3} (2 + 3x) dx can be split into two integrals: \int{1}^{3} 2 dx + \int_{1}^{3} 3x dx.
The integral \int_{1}^{3} 2 dx represents the area of a rectangle with height 2 and width 3 - 1 = 2, so the area is 2 \cdot 2 = 4.
The integral \int_{1}^{3} 3x dx represents the area under the line 3x, which forms a trapezoid.
At x = 1, 3x = 3, and at x = 3, 3x = 9.
The area of this trapezoid is \frac{1}{2}(3 + 9) \cdot 2 = 12.
The sum of the two areas is 4 + 12 = 16.

Switching the limits of integration changes the sign: \int{a}^{b} f(x) dx = -\int{b}^{a} f(x) dx.
Moving in the opposite direction negates the integral.
Breaking up the interval of integration: \int{a}^{b} f(x) dx = \int{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx.
The integral from a to b can be broken into two integrals by picking some point C in between.
The integral of a sum (or difference) is the sum (or difference) of the integrals: \int{a}^{b} [f(x) \pm g(x)] dx = \int{a}^{b} f(x) dx \pm \int_{a}^{b} g(x) dx.
A constant factor can be pulled out of the integral: \int{a}^{b} c \cdot f(x) dx = c \cdot \int{a}^{b} f(x) dx.

Given \int{0}^{1} f(t) dt = 1 and \int{2}^{1} f(t) dt = 3, evaluate \int_{0}^{2} [3f(t) - 1] dt.
First, break up the integral: \int{0}^{2} 3f(t) dt - \int{0}^{2} 1 dt.
Pull out the constant: 3 \int{0}^{2} f(t) dt - \int{0}^{2} 1 dt.
Evaluate the integral of 1 from 0 to 2: \int_{0}^{2} 1 dt = 2.
Rewrite the integral from 0 to 2 as \int{0}^{1} f(t) dt + \int{1}^{2} f(t) dt.
Use the property \int{1}^{2} f(t) dt = -\int{2}^{1} f(t) dt = -3.
So, \int_{0}^{2} f(t) dt = 1 - 3 = -2.
Plug back into the original expression: 3(-2) - 2 = -8.

Given \int{0}^{\frac{\pi}{4}} \tan(x) dx = \frac{\ln(2)}{2}, evaluate \int{0}^{\frac{\pi}{4}} [1 + 4\tan(x)] dx.
Break up the integral: \int{0}^{\frac{\pi}{4}} 1 dx + \int{0}^{\frac{\pi}{4}} 4\tan(x) dx.
Pull out the constant: \int{0}^{\frac{\pi}{4}} 1 dx + 4 \int{0}^{\frac{\pi}{4}} \tan(x) dx.
Evaluate the integral of 1 from 0 to \frac{\pi}{4}: \int_{0}^{\frac{\pi}{4}} 1 dx = \frac{\pi}{4}.
Substitute the given value: \frac{\pi}{4} + 4 \cdot \frac{\ln(2)}{2} = \frac{\pi}{4} + 2\ln(2).