Loading [MathJax]/extensions/mml2jax.js
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

Search

  • Filter Results
  • Location
  • Classification
    • Article type
    • Stage
    • Author
    • Cover Page
    • License
    • Show Page TOC
    • Transcluded
    • PrintOptions
    • OER program or Publisher
    • Autonumber Section Headings
    • License Version
    • Print CSS
    • Screen CSS
    • Number of Print Columns
  • Include attachments
Searching in
About 3 results
  • 4.5: Double Integrals Over General Regions
    https://math.libretexts.org/Courses/Misericordia_University/MTH_226%3A_Calculus_III/Chapter_15%3A_Multiple_Integration/4.05%3A_Double_Integrals_Over_General_Regions
    \[\begin{align*} \int_{x=0}^{x=2}\int_{y=\frac{1}{2}x}^{y=1}x^2e^{xy}\,dy\,dx &= \int_{x=0}^{x=2}\left[\int_{y=\frac{1}{2}x}^{y=1}x^2e^{xy}\,dy\right] dx &\text{Iterated integral for a Type I region.}...\[\begin{align*} \int_{x=0}^{x=2}\int_{y=\frac{1}{2}x}^{y=1}x^2e^{xy}\,dy\,dx &= \int_{x=0}^{x=2}\left[\int_{y=\frac{1}{2}x}^{y=1}x^2e^{xy}\,dy\right] dx &\text{Iterated integral for a Type I region.}\\ &=\int_{x=0}^{x=2} \left.\left[ x^2 \frac{e^{xy}}{x} \right] \right|_{y=1/2x}^{y=1}\,dx & \text{Integrate with respect to $y$}\\ &= \int_{x=0}^{x=2} \left[xe^x - xe^{x^2/2}\right]dx &\text{Integrate with respect to $x$} \\ &=\left[xe^x - e^x - e^{\frac{1}{2}x^2} \right] \Big|_{x=0}^{x=2} = 2. \e…
  • Double Integrals Over General Regions
    https://math.libretexts.org/Courses/Montana_State_University/M273%3A_Multivariable_Calculus/15%3A_Multiple_Integration/Double_Integrals_Over_General_Regions
    \[\begin{align*} \int_{x=0}^{x=2}\int_{y=\frac{1}{2}x}^{y=1}x^2e^{xy}\,dy\,dx &= \int_{x=0}^{x=2}\left[\int_{y=\frac{1}{2}x}^{y=1}x^2e^{xy}\,dy\right] dx &\text{Iterated integral for a Type I region.}...\[\begin{align*} \int_{x=0}^{x=2}\int_{y=\frac{1}{2}x}^{y=1}x^2e^{xy}\,dy\,dx &= \int_{x=0}^{x=2}\left[\int_{y=\frac{1}{2}x}^{y=1}x^2e^{xy}\,dy\right] dx &\text{Iterated integral for a Type I region.}\\ &=\int_{x=0}^{x=2} \left.\left[ x^2 \frac{e^{xy}}{x} \right] \right|_{y=1/2x}^{y=1}\,dx & \text{Integrate with respect to $y$}\\ &= \int_{x=0}^{x=2} \left[xe^x - xe^{x^2/2}\right]dx &\text{Integrate with respect to $x$} \\ &=\left[xe^x - e^x - e^{\frac{1}{2}x^2} \right] \Big|_{x=0}^{x=2} = 2. \e…
  • Double Integrals Over General Regions
    https://math.libretexts.org/Courses/Georgia_State_University_-_Perimeter_College/MATH_2215%3A_Calculus_III/15%3A_Multiple_Integration/Double_Integrals_Over_General_Regions
    \[\begin{align*} \int_{x=0}^{x=2}\int_{y=\frac{1}{2}x}^{y=1}x^2e^{xy}\,dy\,dx &= \int_{x=0}^{x=2}\left[\int_{y=\frac{1}{2}x}^{y=1}x^2e^{xy}\,dy\right] dx &\text{Iterated integral for a Type I region.}...\[\begin{align*} \int_{x=0}^{x=2}\int_{y=\frac{1}{2}x}^{y=1}x^2e^{xy}\,dy\,dx &= \int_{x=0}^{x=2}\left[\int_{y=\frac{1}{2}x}^{y=1}x^2e^{xy}\,dy\right] dx &\text{Iterated integral for a Type I region.}\\ &=\int_{x=0}^{x=2} \left.\left[ x^2 \frac{e^{xy}}{x} \right] \right|_{y=1/2x}^{y=1}\,dx & \text{Integrate with respect to $y$}\\ &= \int_{x=0}^{x=2} \left[xe^x - xe^{x^2/2}\right]dx &\text{Integrate with respect to $x$} \\ &=\left[xe^x - e^x - e^{\frac{1}{2}x^2} \right] \Big|_{x=0}^{x=2} = 2. \e…

Support Center

How can we help?