Introduction

**What it means:** a definite integral evaluates to a value; this value is the area under the curve of a function from points $x=a$ to $x=b$.

No constant of integration, thus "definite". Also, the integral symbol is an elongated "s", which stands for *summa*, or sum.

Written as $\displaystyle\int_a^bf(x)dx$

Integral Properties

$\displaystyle\int_a^bf(x)dx=-\displaystyle\int_b^af(x)dx$

$\displaystyle\int_a^af(x)dx=0$

$\displaystyle\int_a^cf(x)dx=\displaystyle\int_a^bf(x)dx+\displaystyle\int_b^cf(x)dx$

$\displaystyle\int_a^bf(x)+g(x)dx=\displaystyle\int_a^bf(x)dx+\displaystyle\int_a^bg(x)dx$

$\displaystyle\int_a^bCf(x)dx=C\displaystyle\int_a^bf(x)dx$

$\displaystyle\int_a^af(x)dx=0$

$\displaystyle\int_a^cf(x)dx=\displaystyle\int_a^bf(x)dx+\displaystyle\int_b^cf(x)dx$

$\displaystyle\int_a^bf(x)+g(x)dx=\displaystyle\int_a^bf(x)dx+\displaystyle\int_a^bg(x)dx$

$\displaystyle\int_a^bCf(x)dx=C\displaystyle\int_a^bf(x)dx$

Fundamental Theorem of Calculus

The First: We can evaluate definite integrals using indefinite integrals:

$\displaystyle\int_a^bf(x)dx=F(x)\big\rvert_a^b=F(b)-F(a)$

You evaluate the indefinite integral, and then evaluate it at $x=a$ and $x=b$, and subtract the values.

If you could take one thing away from any calculus course, **this is it**.

The Second (aka the corollary to the FTC aka Leibniz Rule): The derivative of an integral with variable endpoint. If $F(x)=\displaystyle\int_a^xf(t)dt$, then $F'(x)=f(x)$. The reason for this is, if we were to integrate $f(t)$ using the First fundamental theorem, we would get $F(x)-F(a)$. Since $a$ is a constant, $F(a)$ is also a constant, and when we take the derivative of anything the constant always drops; thus, we are left with $f(x)$: $F'(x)-F'(a)=f(x)-0=f(x)$.

What happens if we have a variable in the lower bound? Or in both? Or what if the variable isn't a variable but a function? What if we wanted to evaluate the integral with one or more endpoints being infinity? The First and Second fundamental theorems complement each other in solving such problems.

Integral with variable in lower bound: Just like we did earlier, we can use the First fundamental theorem to evaluate this. Doing so, we get $\frac{d}{dx}\displaystyle\int^a_xf(t)dt=F'(a)-F'(x)=0-f(x)=-f(x)$. Another way of looking at this is realizing that $\displaystyle\int_x^af(t)dt=-\displaystyle\int^x_af(t)dt$.

Integral with function in one bound: If, say, we have an integral that looks like this $F(x)=\displaystyle\int_a^{g(x)}f(x)dx$. We can further rewrite this as the composition of two functions. If we say $h(x)=\displaystyle\int_a^xf(x)dx$, then we can write $F(x)=h(g(x))$. If we wanted to take the derivative of this, we would have to implement the chain rule: $F'(x)=h'(g(x))g'(x)$. We can then treat this exactly the same as with regular variable bounds. Basically, if you have a function for a bound, you need to use the corollary the FTC to substitute the function into the integrand, then multiply the result by the derivative of the function.

Integral with function in both bounds: We know that if we have $\displaystyle\int^{g(x)}_{h(x)}f(t)dt$, that this can be broken down into $\displaystyle\int_{h(x)}^cf(t)dt+\displaystyle\int_c^{g(x)}f(t)dt$, where $c$ is some arbitrary point lying between $h(x)$ and $g(x)$. From here we can evaluate the integrals using the strategies mentioned before.

Improper integrals: Improper integrals are integrals which have $\pm \infty$ as one or both of their endpoints. Since infinity is NOT a number, you can't directly plug it in using the first fundamental theorem, so you will have to get creative. Remember how all of Calculus is built on limits? You can rewrite $\displaystyle\int_a^{\infty}f(x)dx$ as $\displaystyle\lim_{t \to \infty}\displaystyle\int_a^tf(x)dx$. Now, this becomes a simple problem; just use the corollary to the FTC and evaluate the limit! Note that if you have an integral from $-\infty$ to $\infty$, you can break it up using the same technique as if you had a function in both bounds.

Methods of Approximation

Trapezoidal Rule

Approximating area under the curve using trapezoids

Subintervals are the equally spaced $x$-values which are the base of the trapezoids

Subinterval size: $\Delta x=\frac{b-a}{n}$, where $n$ is the number of subintervals

Formula is $\displaystyle\int_a^bf(x)dx\approx\frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+...+f(x_n)]$

This is just finding the area of the individual trapezoids and summing them

Rule of thumb: the smaller the subinterval, the better the approximation

Simpson's Rule
Approximating area under curve using quadratics

For this to work, the number of subintervals must be *even* and you need at least three points

Formula is

$\displaystyle\int_a^bf(x)dx\approx\frac{\Delta x}{3}[f(x_0)+4f(x_1)+2f(x_2)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)]$

Alternates coefficients 1,4,2,4,2,...2,4,1

Riemann Sum
Approximation with sum of rectangles, the more rectangles and smaller the subintervals, the more accurate the estimate

Five different types:

Left: Height of the rectangle is the leftmost point on the subinterval

Right: Height of the rectangle is the rightmost point on the subinterval

Upper: Height of the rectangle is the uppermost point on the subinterval

Lower: Height of the rectangle is the lowerpost point on the subinterval

Midpoint: Height of the rectangle is the midpoint of the subinterval

Right: Height of the rectangle is the rightmost point on the subinterval

Upper: Height of the rectangle is the uppermost point on the subinterval

Lower: Height of the rectangle is the lowerpost point on the subinterval

Midpoint: Height of the rectangle is the midpoint of the subinterval

Different sums result in over or underestimates depending on the function

The Riemann Sum leads into the formal definition of the definite integral: $\displaystyle\int_a^bf(x)dx=\lim\limits_{n \to \infty}{\displaystyle\sum_{i=1}^n f(x_i)\Delta x}$

The smaller the $\Delta x$, the more accurate the approximation.

$\displaystyle\sum_{i=1}^{\infty}$ is shorthand for $\lim\limits_{n \to \infty}{\displaystyle\sum_{i=1}^n}$

$x_i=a+(\Delta x)i$. You should already be familiar with Summation Notation, and should be comfortable with converting between Riemann integral form and standard integral form. To solve the integral in Riemann sum form, simply move all values containing $n$ outside of the sum and evaluate.

Some sum properties:

$\displaystyle\sum_{i=1}^n c=cn$

$\displaystyle\sum_{i=1}^{n} i=\frac{n(n+1)}{2}$

$\displaystyle\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}$

$\displaystyle\sum_{i=1}^n i^3= \left(\frac{n(n+1)}{2}\right)^2$

$\displaystyle\sum_{i=1}^{n} i=\frac{n(n+1)}{2}$

$\displaystyle\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}$

$\displaystyle\sum_{i=1}^n i^3= \left(\frac{n(n+1)}{2}\right)^2$