Evaluate [latex]\int_{y=0}^{y=1}\int_{x=0}^{x=2-2y}4x+5\,\mathrm{d}x\,\mathrm{d}y[/latex] over the region described in Example 27.11 .

We first perform the inner integral [latex]\int_{x=0}^{x=2-2y}4x+5\,\mathrm{d}x[/latex] integrating with respect to x. This gives [latex]\begin{array}{c}{{[2x^{2}+5x]_{x=0}^{x=2-2y}=2(2-2y)^{2}+5(2-2y)-0}}\\ {{{}=8y^{2}-26y+18}}\end{array}[/latex] Next we perform the outer integral [latex]\int_{y=0}^{y=1}8y^{2}-26y+18\,\mathrm{d}y=\left[{\frac{8y^{3}}{3}}-{\frac{26y^{2}}{2}}+18y\right]_{0}^{1}\\[/latex] [latex]={\frac{8}{3}}-{\frac{26}{2}}+18[/latex] = 7.667 The region of integration is shown in Figure 27.12. The first integral, with respect to x, corresponds to integrating along the horizontal strip from x = 0 to x = 2 − 2y. Then as y varies from 0 to 1 in the second integral, the horizontal strips will cover the entire region.

Question 27.12: Evaluate ∫y = 0 ˆ y = 1 ∫x = 0 ˆ x = 2 - 2y 4x+5dx dy over t......

Engineering Mathematics A foundation for Electronic, Electrical, Communications and Systems Engineers [1040491]

Evaluate

\int_{y=0}^{y=1}\int_{x=0}^{x=2-2y}4x+5\,\mathrm{d}x\,\mathrm{d}y

over the region described in Example 27.11.

Step-by-Step

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We first perform the inner integral

\int_{x=0}^{x=2-2y}4x+5\,\mathrm{d}x

integrating with respect to x. This gives

\begin{array}{c}{{[2x^{2}+5x]_{x=0}^{x=2-2y}=2(2-2y)^{2}+5(2-2y)-0}}\\ {{{}=8y^{2}-26y+18}}\end{array}

Next we perform the outer integral

\int_{y=0}^{y=1}8y^{2}-26y+18\,\mathrm{d}y=\left[{\frac{8y^{3}}{3}}-{\frac{26y^{2}}{2}}+18y\right]_{0}^{1}\\

={\frac{8}{3}}-{\frac{26}{2}}+18

= 7.667

The region of integration is shown in Figure 27.12. The first integral, with respect to x, corresponds to integrating along the horizontal strip from x = 0 to x = 2 − 2y. Then as y varies from 0 to 1 in the second integral, the horizontal strips will cover the entire region.