#### Related to [[Integration]] --- ## Type 1 improper integral - A type 1 improper integral is any integral whose **upper/lower bound is equal to infinity**. - To solve these integrals we must first **evaluate the integrand, finding the antiderivative**. Then we evaluate the **infinite bound** using a [[Limit]] where x approaches either positive or negative infinity. If one of the bounds is a real integer than we say that the **integral converges**. Otherwise we say the integral **diverges**. $\newline$ $\text{Improper integral type 1:}$ $\text{Type 1 improper =}\int_a^bf(x)\,dx\;\text{ where a or b}=\infty$ $\color{LightCyan}\int_{-\infty}^1f(x)\,dx\longrightarrow F(1)-\biggl(\lim_{t\rightarrow\infty}F(t)\biggl)$$\newline$ - If both the **upper and lower bound** are equal to infinity we must **split the integral up into two separate integrals** and evaluate them separately. One will go from **negative infinity to 0** and the other will go from **0 to positive infinity**. - *Note: If one of these integrals diverges than we say that the integral **overall** also diverges.* $\newline$ $\text{Transform:}\int_{-\infty}^{\infty}f(x)\,dx\;\text{ into: }\int_{-\infty}^0f(x)\,dx+\int_0^{\infty}f(x)\,dx$ $\newline$ --- ## Type 2 improper integral - A type two improper integral is an integral that contains **a vertical asymptote somewhere on the bounds of integration**. - NOTE: If the integral is *in terms of y* rather than, x then we would say that it is type 2 improper if it contains a **horizontal asymptote** somewhere on its bounds. - If the vertical asymptote **is somewhere** ==**between the upper and lower bounds**== we must split the integral up at the **point of this asymptote**. This is because we **can only evaluate type 2 improper integrals if the vertical discontinuity (asymptote) has the same coordinate as either the upper or lower bound**. $\newline$ $\text{If the asymptote is at }x=c\text{ and }a<c<b\;\int_a^b f(x)\,dx\longrightarrow\int_a^cf(x)\,dx+\int_c^bf(x)\,dx$ $\newline$ - To evaluate type 2 improper integrals we must use a [[Limit]] that approaches the x *(or y)* value of the asymptote. If the asymptote's location is **equal to the integral's lower bound** the limit will approach its value from the **positive side**. If the asymptote's location is equal to the upper bound then the limit must approach the asymptote's value from the negative side. $\newline$ $\int_a^bf(x)\,dx\;{\color{grey}\small\text{ where the asymptote is at }x=b}\longrightarrow\lim_{t\rightarrow b^-}\Big(F(t)\Big)-F(a)$ $\newline$ >[!check] >##### *Remember:* >- In a type 2 improper integral we the asymptote must be exactly at either the **upper or lower bound**. If it is not we must split up the integral **at the point of the asymptote** in order to ensure that it is at the upper bound of one integral and lower bound of the other. --- ## Convergence and divergence - We say that an integral **converges if its area is equal to positive or negative infinity**. We say that an integral diverges if **its area is equal to some finite real number**. - Both type 1 and type 2 improper integrals **diverge** if they have an **infinite area** and converge if they have **a finite area**. $\newline$ $\text{For example: }\int_{1}^{\infty}\frac{1}{x^2}\,dx\;\text{ converges and }\int_{1}^{\infty}\frac{1}{x}\,dx\;\text{ diverges}$ $\newline$ >[!tip] >##### *Useful rule:* >$\int_{1}^{\infty}\frac{1}{x^p}\,dx\small\;\text{is convergent if }p>1\text{ and divergent if }p\le 1$ - The **comparison theorem** allows us to determine if an integral converges or diverges by comparing it to *another integral* that is either larger or smaller everywhere on the bounds. - For example, say that our function is $f(x)$ and $g(x)$ is another function that is **smaller than $f(x)$ on the bounds of integration.** The comparison theorem says that **if g(x) diverges than $f(x)$ must also diverge**. This makes sense because, if $f(x)$ is truly larger on the bounds of integration than $g(x)$, it's integral will be bounded the smaller function. - The comparison theorem also works if $g(x)$ is larger on the bounds than $f(x)$. If this is the case **the comparison theorem states that: if $g(x)$ converges than $f(x)$ must also converge** which again makes sense because $f(x)$ is bounded below $g(x)$. $\newline$ $\large\text{The comparison theorem:}$ $\text{let }f(x)\ge g(x)\ge 0\;\text{ for}\;x\ge a$ $\newline$ $\text{If }\int_a^{\infty}f(x)\,dx\,\text{\color{grey} is convergent than }\int_a^{\infty}g(x)\,dx\,\text{\color{grey} is convergent}$ $\text{If }\int_a^{\infty}f(x)\,dx\,\text{\color{grey} is divergent than }\int_a^{\infty}g(x)\,dx\,\text{\color{grey} is divergnet}$ >[!quote] >##### *Visual of the comparison theorem:* >![[Comparisontheoremvisual.png|850]] --- ## Calculator >[!cite] >--- ><iframe src="https://www.wolframalpha.com/input/?i=improper+integral+calculator" height="1000" width="878"></iframe>