# + C Definite Integrals The fundamental Theorem of Calculus Let f be continuous on [a,b]. Then: âˆ«[f(x) dx] = F(b) – F(a) where F is any antiderivative of f, i.e., F'(x) = f(x) Area between two curves

Calculus – Integration Basics (1 of 5) Integrating a Constant
Calculus – Integration Basics (1 of 5) Integrating a Constant

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Basic Rules of Integration
Indefinite integral of a constant: âˆ«fk du = ku + C
Power rule: âˆ«Ju^” du = (n + 1)u^(n+1) + C
Constant multiple rule: âˆ«Jrf(u) du = kâˆ«f(u) du where k is a constant
Sum rule: âˆ«[f(u) + g(u)] du = âˆ«f(u) du + âˆ«g(u) du
Exponential function: âˆ«e^u du = e^u + C
Logarithmic function: âˆ«1/u du = ln|u| + C
Definite Integrals
The fundamental Theorem of Calculus
Let f be continuous on [a,b]. Then: âˆ«[f(x) dx] = F(b) – F(a) where F is any antiderivative of f, i.e., F'(x) = f(x)
Area between two curves
Let f and g be continuous functions such that f(x) â‰¥ g(x) on the interval [a,b]. Then the area of the region bounded by [a,b] is given by âˆ«[f(x) – g(x)] dx
Calculus of Several Variables
Determining Relative Extrema
D(x,y) = f(x,y) – f(x,y)
If D > 0 and f_x = 0, a relative minimum point occurs at (x,y)
If D > 0 and f_x = 0, a relative maximum point occurs at (x,y)
If D < 0, (x,y) is neither a maximum nor minimum
If D = 0, the test is inconclusive

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01:17

Integration by parts leads to a rule for integrating inverses that usuallygives good results:\begin{aligned} \int f^{-1}(x) d x &=\int y f^{\prime}(y) d y \\ &=y f(y)-\int f(y) d y \\ &=x f^{-1}(x)-\int f(y) d y \end{aligned}The idea is to take the most complicated part of the integral, in this case$f^{-1}(x),$ and simplify it first. For the integral of $\ln x,$ we get\begin{aligned} \int \ln x d x &=\int y e^{y} d y \\ &=y e^{y}-e^{y}+C \\ &=x \ln x-x+C \end{aligned}For the integral of $\cos ^{-1} x$ we get\begin{aligned} \int \cos ^{-1} x d x &=x \cos ^{-1} x-\int \cos y d y \\ &=x \cos ^{-1} x-\sin y+C \\ &=x \cos ^{-1} x-\sin \left(\cos ^{-1} x\right)+C \end{aligned}Use the formula$$\int f^{-1}(x) d x=x f^{-1}(x)-\int f(y) d y$$to evaluate the integrals. Express your answers in terms of $x .$$\int \sec ^{-1} x d x$$Another way to integrate$f^{-1}(x)$(when$f^{-1}$is integrable) is touse integration by parts with$u=f^{-…

01:02

Integration by parts leads to a rule for integrating inverses that usuallygives good results:\begin{aligned} \int f^{-1}(x) d x &=\int y f^{\prime}(y) d y \\ &=y f(y)-\int f(y) d y \\ &=x f^{-1}(x)-\int f(y) d y \end{aligned}The idea is to take the most complicated part of the integral, in this case$f^{-1}(x),$ and simplify it first. For the integral of $\ln x,$ we get\begin{aligned} \int \ln x d x &=\int y e^{y} d y \\ &=y e^{y}-e^{y}+C \\ &=x \ln x-x+C \end{aligned}For the integral of $\cos ^{-1} x$ we get\begin{aligned} \int \cos ^{-1} x d x &=x \cos ^{-1} x-\int \cos y d y \\ &=x \cos ^{-1} x-\sin y+C \\ &=x \cos ^{-1} x-\sin \left(\cos ^{-1} x\right)+C \end{aligned}Use the formula$$\int f^{-1}(x) d x=x f^{-1}(x)-\int f(y) d y$$to evaluate the integrals. Express your answers in terms of $x .$$\int \sin ^{-1} x d x$$Another way to integrate$f^{-1}(x)$(when$f^{-1}$is integrable) is touse integration by parts with$u=f^{-…

02:31

Integration by parts leads to a rule for integrating inverses that usuallygives good results:\begin{aligned} \int f^{-1}(x) d x &=\int y f^{\prime}(y) d y \\ &=y f(y)-\int f(y) d y \\ &=x f^{-1}(x)-\int f(y) d y \end{aligned}The idea is to take the most complicated part of the integral, in this case$f^{-1}(x),$ and simplify it first. For the integral of $\ln x,$ we get\begin{aligned} \int \ln x d x &=\int y e^{y} d y \\ &=y e^{y}-e^{y}+C \\ &=x \ln x-x+C \end{aligned}For the integral of $\cos ^{-1} x$ we get\begin{aligned} \int \cos ^{-1} x d x &=x \cos ^{-1} x-\int \cos y d y \\ &=x \cos ^{-1} x-\sin y+C \\ &=x \cos ^{-1} x-\sin \left(\cos ^{-1} x\right)+C \end{aligned}Use the formula$$\int f^{-1}(x) d x=x f^{-1}(x)-\int f(y) d y$$to evaluate the integrals. Express your answers in terms of $x .$$\int \tan ^{-1} x d x$$Another way to integrate$f^{-1}(x)$(when$f^{-1}$is integrable) is touse integration by parts with$u=f^{-…

01:53

Consider the solid under the graph of z = e −x 2−y 2 above the disk x 2 +y 2 ≤ a 2 , where a > 0. (a) Set up the integral to find the volume of the solid. Instructions: Please enter the integrand in the first answer box, typing theta for θ. Depending on the order of integration you choose, enter dr and dtheta in either order into the second and third answer boxes with only one dr or dtheta in each box. Then, enter the limits of integration. 14 va

01:22

calculus

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