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Find int_(0)^(2)f(x,y)dx and int_(0)^(3)f(x,y)dy. {:[f(x","y)=11 ysqrt(x+2)],[int_(0)^(2)f(x","y)dx=◻],[int_(0)^(3)f(x","y)dy=◻]:}

Find ∫02f(x,y)dx \int_{0}^{2} f(x, y) d x and ∫03f(x,y)dy \int_{0}^{3} f(x, y) d y .\newlinef(x,y)=11yx+2∫02f(x,y)dx=□∫03f(x,y)dy=□ \begin{array}{r} f(x, y)=11 y \sqrt{x+2} \\ \int_{0}^{2} f(x, y) d x=\square \\ \int_{0}^{3} f(x, y) d y=\square \end{array}

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Q. Find ∫02f(x,y)dx \int_{0}^{2} f(x, y) d x and ∫03f(x,y)dy \int_{0}^{3} f(x, y) d y .\newlinef(x,y)=11yx+2∫02f(x,y)dx=□∫03f(x,y)dy=□ \begin{array}{r} f(x, y)=11 y \sqrt{x+2} \\ \int_{0}^{2} f(x, y) d x=\square \\ \int_{0}^{3} f(x, y) d y=\square \end{array}
  1. Given Function and Integral: We are given the function f(x,y)=11yx+2f(x, y) = 11y\sqrt{x+2}. We need to find the integral of this function with respect to xx from 00 to 22. Let's denote this integral as IxI_x.\newlineIx=∫0211yx+2 dxI_x = \int_{0}^{2} 11y\sqrt{x+2} \, dx\newlineSince yy is treated as a constant with respect to xx, we can pull it out of the integral.\newlineIx=11y∫02x+2 dxI_x = 11y \int_{0}^{2} \sqrt{x+2} \, dx\newlineNow we need to find the antiderivative of x+2\sqrt{x+2} with respect to xx.\newlineLet xx11, then xx22.\newlineThe integral becomes:\newlinexx33\newlineThe antiderivative of xx44 is xx55.\newlineSo, xx66 evaluated from xx77 to xx88.
  2. Pulling Out Constants: Now we evaluate the antiderivative at the upper and lower limits of integration. \newlineIx=11y×23×[432−232]I_x = 11y \times \frac{2}{3} \times [4^{\frac{3}{2}} - 2^{\frac{3}{2}}]\newline432=(22)32=23=84^{\frac{3}{2}} = (2^2)^{\frac{3}{2}} = 2^3 = 8\newline232=23=8=222^{\frac{3}{2}} = \sqrt{2^3} = \sqrt{8} = 2\sqrt{2}\newlineSo, Ix=11y×23×[8−22]I_x = 11y \times \frac{2}{3} \times [8 - 2\sqrt{2}]\newlineIx=11y×23×[8−22]I_x = 11y \times \frac{2}{3} \times [8 - 2\sqrt{2}]\newlineIx=223y×[4−2]I_x = \frac{22}{3}y \times [4 - \sqrt{2}]
  3. Finding Antiderivative of x+2\sqrt{x+2}: Next, we need to find the integral of the function f(x,y)f(x, y) with respect to yy from 00 to 33. Let's denote this integral as IyI_y.
    Iy=∫0311yx+2 dyI_y = \int_{0}^{3} 11y\sqrt{x+2} \, dy
    Since xx is treated as a constant with respect to yy, we can pull x+2\sqrt{x+2} out of the integral.
    f(x,y)f(x, y)00
    Now we need to find the antiderivative of f(x,y)f(x, y)11 with respect to yy.
    The antiderivative of f(x,y)f(x, y)11 is f(x,y)f(x, y)44.
    So, f(x,y)f(x, y)55 evaluated from f(x,y)f(x, y)66 to f(x,y)f(x, y)77.
  4. Evaluating Antiderivative at Limits: Now we evaluate the antiderivative at the upper and lower limits of integration.\newlineIy=x+2×(112)×[32−02]I_y = \sqrt{x+2} \times \left(\frac{11}{2}\right) \times [3^2 - 0^2]\newline32=93^2 = 9\newlineSo, Iy=x+2×(112)×9I_y = \sqrt{x+2} \times \left(\frac{11}{2}\right) \times 9\newlineIy=x+2×(992)I_y = \sqrt{x+2} \times \left(\frac{99}{2}\right)

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