evaluate the double integral (iint_{r}(x^{2}+4y)da), where the regoin (r) is bounded by the lines (y = x)…

Explanation:

Step1: Determine the limits of integration

The curves (y = x) and (y=x^{3}) intersect at (x = 0) and (x = 1) (since (x=x^{3}\Rightarrow x^{3}-x=0\Rightarrow x(x - 1)(x + 1)=0), and considering the region with (x\geq0)). For a given (x\in[0,1]), the lower - limit of (y) is (y = x^{3}) and the upper - limit is (y=x). The double integral (\iint_{R}(x^{2}+4y)dA=\int_{0}^{1}\int_{x^{3}}^{x}(x^{2}+4y)dydx)

Step2: Integrate with respect to (y)

First, integrate (\int_{x^{3}}^{x}(x^{2}+4y)dy). Using the power rule (\int(ax^{n}+bx^{m})dx=\frac{ax^{n + 1}}{n+1}+\frac{bx^{m+1}}{m + 1}+C) ((n\neq-1,m\neq-1)), we have: (\int_{x^{3}}^{x}(x^{2}+4y)dy=\left[x^{2}y+4\times\frac{y^{2}}{2}\right]_{y = x^{3}}^{y = x}) (=x^{2}(x - x^{3})+2(x^{2}-x^{6})) (=x^{3}-x^{5}+2x^{2}-2x^{6})

Step3: Integrate with respect to (x)

Now, integrate (\int_{0}^{1}(x^{3}-x^{5}+2x^{2}-2x^{6})dx) (\int_{0}^{1}(x^{3}-x^{5}+2x^{2}-2x^{6})dx=\left[\frac{x^{4}}{4}-\frac{x^{6}}{6}+2\times\frac{x^{3}}{3}-2\times\frac{x^{7}}{7}\right]_{0}^{1}) (=\frac{1}{4}-\frac{1}{6}+\frac{2}{3}-\frac{2}{7}) Find a common denominator (the common denominator of (4,6,3,7) is (84)): (\frac{1\times21}{4\times21}-\frac{1\times14}{6\times14}+\frac{2\times28}{3\times28}-\frac{2\times12}{7\times12}) (=\frac{21-14 + 56-24}{84}=\frac{39}{84}=\frac{13}{28})

Answer:

(\frac{13}{28})