6.1 Areas Between Curves/12: Difference between revisions

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&= \left[\frac{4x^2}{2}\right]\Bigg|_{0}^{2} \\[2ex]





Revision as of 23:18, 22 September 2022

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} & \color{red}\mathbf{y=x^2} & \color{royalblue}\mathbf{y=4x-x^2} \\ & x=0 & x=2 \\ \end{align} }


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{2} \left[(4x-x^2) - (x^2)\right] dx}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} 4-x^2 &= x^2 \\ 4x-2x^2 &= 0 \\ 2x(2-x) &= 0 \\ x &= 0& x = 2 \end{align} }


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{2} \left[(4x-x^2) - (x^2)\right]dx = \int_{0}^{2}(4x)dx &= \left[\frac{-16}{3}+16\right]-\left[\frac{-54}{3}+24\right] = \frac{38}{3}-8 \\[2ex] &= \frac{14}{3} \end{align} }


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_{-2}^{2} \left((8-x^2) - (x^2)\right)dx &= \int_{-2}^{2}\left(8-2x^2\right)dx \\[2ex] &= \left[8x-\frac{2x^3}{3}\right]\Bigg|_{-2}^{2} \\[2ex] &= \left[8(2)-\frac{2(2)^3}{3}\right] - \left[8(-2)-\frac{2(-2)^3}{3}\right] \\[2ex] &= \left[16-\frac{16}{3}\right]-\left[-16+\frac{16}{3}\right] = 32-\frac{32}{3} \\[2ex] &= \frac{64}{3} \end{align} }