Answer to Question #89653 in Quantitative Methods for lana majeed

Question #89653
Evaluate following double integral value taking 4 segment for both x and y direction use Simpson rule ∫(2)_(4)∫(x)_(3x^2) ((3x^2)+y) dy dx
1
Expert's answer
2019-05-15T14:15:52-0400
"\\displaystyle\\int_{2}^4\\displaystyle\\int_{x}^{3x^2}(3x^2+y)dydx"

 We use the Composite Simpson’s rule to approximate the “inner” integral.


"F(x)=\\displaystyle\\int_{x}^{3x^2}(3x^2+y)dy\\approx""\\approx{k(x) \\over 3}\\bigg[f\\Big(x, c(x)\\Big)+4f\\Big(x,c(x)+k(x)\\Big)+""+2f\\Big(x, c(x)+2k(x)\\Big)+4f\\Big(x,c(x)+3k(x)\\Big)+""+f\\Big(x, d(x)\\Big)\\bigg]"

"f(x, y)=3x^2+y""c(x)=x,\\ d(x)=3x^2,\\ k(x)={d(x)-c(x) \\over 4}={3x^2-x \\over 4}"

"f\\Big(x, x\\Big)=3x^2+x""4f\\Big(x, x+{3x^2-x \\over 4}\\Big)=12x^2+4x+3x^2-x=15x^2+3x""2f\\Big(x, x+{3x^2-x \\over 2}\\Big)=6x^2+2x+3x^2-x=9x^2+x""4f\\Big(x, x+3\\cdot{3x^2-x \\over 4}\\Big)=12x^2+4x+9x^2-3x=21x^2+x""f\\Big(x, 3x^2\\Big)=3x^2+3x^2=6x^2"

"\\displaystyle\\int_{x}^{3x^2}(3x^2+y)dy\\approx"

"\\approx{3x^2-x \\over 12}\\bigg[3x^2+x+15x^2+3x+9x^2+x+21x^2+x+6x^2\\bigg]="

"={27x^4-6x^3-x^2 \\over 2}"

Now we may approximate the outer integral.


"\\displaystyle\\int_{2}^4\\displaystyle\\int_{x}^{3x^2}(3x^2+y)dydx\\approx\\displaystyle\\int_{2}^4{27x^4-6x^3-x^2 \\over 2}dx\\approx"


"\\approx{h \\over 3}\\bigg[g(2)+4g(2+h)+2g(2+2h)+4g(2+3h)+g(4)\\bigg]"

"g(x)={27x^4-6x^3-x^2 \\over 2}"

"h={4-2 \\over 4}={1 \\over 2}"

"g(2)={27(2)^4-6(2)^3-(2)^2 \\over 2}=190"

"4g(2+h)=4g({5 \\over 2})=4\\cdot{27({5 \\over 2})^4-6({5 \\over 2})^3-({5 \\over 2})^2 \\over 2}=1909.375"

"2g(2+2h)=2g(3)=2\\cdot{27(3)^4-6(3)^3-(3)^2 \\over 2}=2016"

"4g(2+3h)=4g({7 \\over 2})=4\\cdot{27({7 \\over 2})^4-6({7 \\over 2})^3-({7 \\over 2})^2 \\over 2}=7564.375"

"g(4)={27(4)^4-6(4)^3-(4)^2 \\over 2}=3256"

"\\displaystyle\\int_{2}^4\\displaystyle\\int_{x}^{3x^2}(3x^2+y)dydx\\approx"

"\\approx{1 \\over 6}\\bigg[190+1909.375+2016+7564.375+3256\\bigg]\\approx"

"\\approx2489.292"


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