Question

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

Answer 1

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

 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.

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

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

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

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

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

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

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

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

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

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

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

$$\approx2489.292$$