Workshop 6
Recall that if $y=f(x),$ the area under the curve over the interval $I = [a,b]$ is
$\displaystyle \int_I f(x)dx = \lim_{n\to \infty} \sum_{i=1}^{n} f(x_i^*)(x_i-x_{i-1})$
where $x_i^* \in [x_i,x_{i-1}]$.
$\displaystyle \int_I f(x)dx = \lim_{n\to \infty} \sum_{i=1}^{n} f(x_i^*)(x_i-x_{i-1})$
Right, Middle, Left Riemann sums
Consider the continuous function $f(x)$ on $[a,b]$ and $F(x)$ is function such that $F'(x)=f(x)$ for every $x\in[a,b],$ then
• We call $F(x)$ an anti-derivative.
• This fact is known as The Fundamental Theorem of Calculus.
$\ds \int_1^3 x^2 \, dx$ $\ds = \left[\dfrac{x^3}{3}+C\right]_1^3$ $\qquad \quad \quad\; F(x) = \dfrac{x^3}{3}+C\;$ 👈
$\qquad \qquad \ds = \underbrace{\left[\dfrac{(3)^3}{3}+C\right]}_{F(3)} - \underbrace{\left[\dfrac{(1)^3}{3}+C\right]}_{F(1)}$
$\qquad \qquad \ds = \dfrac{27}{3}+C $ $ -\, \dfrac{1}{3}-C$
$\qquad \qquad \ds = \dfrac{26}{3}$
$\ds \int_1^3 x^2 \, dx$ $\ds = \left[\dfrac{x^3}{3}\right]_1^3$ $\qquad \quad \quad\; F(x) = \dfrac{x^3}{3}\;$ 👈
$\qquad \qquad \ds = \underbrace{\left[\dfrac{(3)^3}{3}\right]}_{F(3)} - \underbrace{\left[\dfrac{(1)^3}{3}\right]}_{F(1)}$
$\qquad \qquad \ds = \dfrac{27}{3} $ $ -\, \dfrac{1}{3}$
$\qquad \qquad \ds = \dfrac{26}{3}$
👉 Note we don't need
to write the constant $C$.
Integration by substitution
\(\ds \int f\left(g\left(x\right)\right)\, dx \) \(\ds = \int f(u)\, du\,\) where \(\,u = g(x).\)
Integration by parts
\(\ds \int \frac{df\left(x\right)}{dx}g\left(x\right)\, dx \) \(\ds = f\left(x\right)g\left(x\right) \) \(\ds \, - \int f\left(x\right)\frac{dg\left(x\right)}{dx}\, du\)
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| Integration by substitution | \(\ds \int f\left(g\left(x\right)\right)\, dx = \int f(u)\, du,\,\) where \(\,u = g(x)\) |
| Integration by parts | \(\ds \int \frac{df\left(x\right)}{dx}g\left(x\right)\, dx = f\left(x\right)g\left(x\right) - \int f\left(x\right)\frac{dg\left(x\right)}{dx}\, du\) |