1. Express as a limit: $\lim_{b \to \infty} \int_0^b e^{-x} dx$. 2) Evaluate the integral: $\lim_{b \to \infty} [-e^{-x}]_0^b = \lim_{b \to \infty} (-e^{-b} + e^0)$. 3) Evaluate the limit: $\lim_{b \to \infty} (-e^{-b} + 1) = 1$. 4) The integral converges to 1.

1. Express as a limit: $\lim_{b \to \infty} \int_1^b \frac{1}{x^2} dx$. 2) Evaluate the integral: $\lim_{b \to \infty} [-\frac{1}{x}]_1^b = \lim_{b \to \infty} (-\frac{1}{b} + 1)$. 3) Evaluate the limit: $\lim_{b \to \infty} (-\frac{1}{b} + 1) = 1$. 4) The integral converges to 1.

1. Express as a limit: $\lim_{a \to 0^+} \int_a^1 \frac{1}{\sqrt{x}} dx$. 2) Evaluate the integral: $\lim_{a \to 0^+} [2\sqrt{x}]_a^1 = \lim_{a \to 0^+} (2 - 2\sqrt{a})$. 3) Evaluate the limit: $\lim_{a \to 0^+} (2 - 2\sqrt{a}) = 2$. 4) The integral converges to 2.

1. Express as a limit: $\lim_{b \to \infty} \int_1^b \frac{1}{x} dx$. 2) Evaluate the integral: $\lim_{b \to \infty} [\ln|x|]_1^b = \lim_{b \to \infty} (\ln(b) - \ln(1))$. 3) Evaluate the limit: $\lim_{b \to \infty} (\ln(b) - 0) = \infty$. 4) The integral diverges.

1. Express as a limit: $\lim_{a \to -\infty} \int_a^0 xe^x dx$. 2) Integrate by parts: $u=x, dv=e^x dx$, so $du=dx, v=e^x$. Then $\int xe^x dx = xe^x - \int e^x dx = xe^x - e^x + C$. 3) Evaluate the integral: $\lim_{a \to -\infty} [xe^x - e^x]_a^0 = \lim_{a \to -\infty} [(0 - e^0) - (ae^a - e^a)] = \lim_{a \to -\infty} [-1 - ae^a + e^a]$. 4) Evaluate the limit: $\lim_{a \to -\infty} [-1 - ae^a + e^a] = -1 - 0 + 0 = -1$. 5) The integral converges to -1.

1. Split the integral: $\int_{-\infty}^{\infty} \frac{1}{1+x^2} dx = \int_{-\infty}^{0} \frac{1}{1+x^2} dx + \int_{0}^{\infty} \frac{1}{1+x^2} dx$. 2) Express as limits: $\lim_{a \to -\infty} \int_a^0 \frac{1}{1+x^2} dx + \lim_{b \to \infty} \int_0^b \frac{1}{1+x^2} dx$. 3) Evaluate the integral: $\lim_{a \to -\infty} [\arctan(x)]_a^0 + \lim_{b \to \infty} [\arctan(x)]_0^b = \lim_{a \to -\infty} [\arctan(0) - \arctan(a)] + \lim_{b \to \infty} [\arctan(b) - \arctan(0)]$. 4) Evaluate the limits: $[0 - (-\frac{\pi}{2})] + [\frac{\pi}{2} - 0] = \frac{\pi}{2} + \frac{\pi}{2} = \pi$. 5) The integral converges to $\pi$.

1. Express as a limit: $\lim_{b \to \infty} \int_2^b \frac{1}{x(x-1)} dx$. 2) Partial fraction decomposition: $\frac{1}{x(x-1)} = \frac{A}{x} + \frac{B}{x-1}$. Solving gives $A = -1$ and $B = 1$. So, $\frac{1}{x(x-1)} = \frac{-1}{x} + \frac{1}{x-1}$. 3) Evaluate the integral: $\lim_{b \to \infty} \int_2^b (\frac{-1}{x} + \frac{1}{x-1}) dx = \lim_{b \to \infty} [-\ln|x| + \ln|x-1|]_2^b = \lim_{b \to \infty} [\ln|\frac{x-1}{x}|]_2^b = \lim_{b \to \infty} [\ln|\frac{b-1}{b}| - \ln|\frac{2-1}{2}|]$. 4) Evaluate the limit: $\lim_{b \to \infty} [\ln|\frac{b-1}{b}| - \ln(\frac{1}{2})] = \ln(1) - \ln(\frac{1}{2}) = 0 - (-\ln(2)) = \ln(2)$. 5) The integral converges to $\ln(2)$.

1. Express as a limit: $\lim_{b \to \infty} \int_0^b cos(x) dx$. 2) Evaluate the integral: $\lim_{b \to \infty} [sin(x)]_0^b = \lim_{b \to \infty} (sin(b) - sin(0))$. 3) Evaluate the limit: $\lim_{b \to \infty} (sin(b) - 0)$. Since $sin(b)$ oscillates between -1 and 1 as b approaches infinity, the limit does not exist. 4) The integral diverges.

1. Split the integral at the discontinuity: $\int_0^{3} \frac{1}{x-2} dx = \int_0^{2} \frac{1}{x-2} dx + \int_2^{3} \frac{1}{x-2} dx$. 2) Express as limits: $\lim_{b \to 2^-} \int_0^b \frac{1}{x-2} dx + \lim_{a \to 2^+} \int_a^3 \frac{1}{x-2} dx$. 3) Evaluate the integral: $\lim_{b \to 2^-} [\ln|x-2|]_0^b + \lim_{a \to 2^+} [\ln|x-2|]_a^3 = \lim_{b \to 2^-} [\ln|b-2| - \ln|-2|] + \lim_{a \to 2^+} [\ln|3-2| - \ln|a-2|] = \lim_{b \to 2^-} [\ln|b-2| - \ln(2)] + \lim_{a \to 2^+} [\ln(1) - \ln|a-2|]$. 4) Evaluate the limits: Since $\lim_{b \to 2^-} \ln|b-2| = -\infty$ and $\lim_{a \to 2^+} \ln|a-2| = -\infty$, both integrals diverge. 5) The integral diverges.

1. Express as a limit: $\lim_{b \to \infty} \int_0^b \frac{x}{(1+x^2)^2} dx$. 2) Use u-substitution: Let $u = 1+x^2$, then $du = 2x dx$, so $x dx = \frac{1}{2} du$. The integral becomes $\frac{1}{2} \int \frac{1}{u^2} du = \frac{1}{2} \int u^{-2} du = \frac{1}{2} [-\frac{1}{u}] + C = -\frac{1}{2(1+x^2)} + C$. 3) Evaluate the integral: $\lim_{b \to \infty} [-\frac{1}{2(1+x^2)}]_0^b = \lim_{b \to \infty} [-\frac{1}{2(1+b^2)} - (-\frac{1}{2(1+0^2)})] = \lim_{b \to \infty} [-\frac{1}{2(1+b^2)} + \frac{1}{2}]$. 4) Evaluate the limit: $\lim_{b \to \infty} [-\frac{1}{2(1+b^2)} + \frac{1}{2}] = 0 + \frac{1}{2} = \frac{1}{2}$. 5) The integral converges to $\frac{1}{2}$.

An integral where the limits of integration involve infinity or the function becomes unbounded within the interval.

The limit of the integral approaches a finite value.

The limit of the integral approaches infinity or does not exist.

The function approaches infinity within the integration interval.

Limits allow us to evaluate integrals with infinite boundaries by approaching infinity.

The function being integrated.

An integral with an upper and lower bound that results in a numerical value.

An integral without boundaries.

A theorem that allows us to solve definite integrals.

Decomposition of a rational function into simpler fractions.

$\int_a^\infty f(x) , dx = \lim_{b \to \infty} \int_a^b f(x) , dx$

$\int_{-\infty}^b f(x) , dx = \lim_{a \to -\infty} \int_a^b f(x) , dx$

$\int_{-\infty}^\infty f(x) , dx = \lim_{a \to -\infty} \int_a^c f(x) , dx + \lim_{b \to \infty} \int_c^b f(x) , dx$

$\int \frac{1}{x} dx = \ln|x| + C$

$\int \frac{1}{a^2+x^2} dx = \frac{1}{a} \arctan(\frac{x}{a}) + C$

$\frac{P(x)}{Q(x)} = \frac{A}{x-a} + \frac{B}{x-b} + ...$

$\int f(g(x))g'(x)dx = \int f(u)du$ where $u=g(x)$

$\int e^x dx = e^x + C$

$\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$

$\int sin(x) dx = -cos(x) + C$