Improper integrals

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  1. Overview of improper integrals
  1. Type 1 integrals with part a
    1. 15xdx \int_{1}^{\infty}\frac{5}{x}dx
    2. 01(4x+5)2dx \int_{0}^{\infty}\frac{1}{(4x+5)^2}dx
  2. Type 1 integrals with part b
    1. 32x4+x2dx \int_{-\infty}^{-3}2x\sqrt{4+x^2}dx
    2. 0exdx \int_{-\infty}^{0}e^xdx
  3. Type 1 integrals with part c
    81+x2dx \int_{-\infty}^{\infty}\frac{8}{1+x^2}dx
    1. Determining convergence and divergence with type 2 integrals
      1. 0πsin2xcosxdx \int_{0}^{\pi}\frac{\sin^2x}{\cos x}dx
      2. 0elnx \int_{0}^{e}\ln x dxdx
      3. 2313xdx \int_{2}^{3}\frac{1}{\sqrt{3-x}}dx
    Topic Notes
    First, we will learn about Type 1 improper integrals. These types of improper integrals have bounds which have positive or negative infinity. Then we will look at Type 2 improper integrals. These improper integrals happen when the function is undefined at a specific place or area within the region of integration. For these integrals, we will have to use limits. If the limit exists and is finite, then the integral can be solved. Otherwise, the integral will be unsolvable.
    There are two types of improper integrals:
    1) Type 1
    a) af(x)dx \int_{a}^{\infty}f(x) dx == lim \limt → \infty atf(x)dx \int_{a}^{t}f(x)dx

    b) bf(x)dx=\int_{-\infty}^{b}f(x)dx=lim\limt →-\inftytbf(x)dx\int_{t}^{b}f(x)dx

    c) f(x)dx=af(x)dx+af(x)dx\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^{a}f(x)dx+\int_{a}^{\infty}f(x)dx

    2) Type 2
    a) If ff is continuous on [a,b)[a,b) and discontinuous at bb, then:
    abf(x)dx=\int_{a}^{b} f(x)dx=lim\limt →b b^-atf(x)dx\int_{a}^{t}f(x)dx

    b) If ff is continuous on (a,b](a,b] and discontinuous at aa, then:
    abf(x)dx=\int_{a}^{b} f(x)dx=lim\limt →a+ a^+tbf(x)dx\int_{t}^{b}f(x)dx

    c) If ff has a discontinuity at cc, where a<c<ba<c<b, then:
    abf(x)dx=acf(x)dx+cbf(x)dx\int_{a}^{b} f(x)dx=\int_{a}^{c} f(x)dx+\int_{c}^{b} f(x)dx

    If the limits exist and is finite, then it is convergent. Otherwise, it is divergent.