Question

Question. $\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\mathit{x}\mathit{d}\mathit{x}\mathit{d}\mathit{y}$, where A is the area between the parabolarole="math" localid="1658895091680" $\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{2}}$ and the straight line 2x-y+8=0.

Short answer

The required solution is 36.

Step-by-step solution

Definition of double integral

The double integral of f(x,y) over the area A in the (x,y) plane as the limit of this sum, and we write it as $\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathit{d}\mathit{x}\mathit{d}\mathit{y}$

Drawing the area bounded by the curve

The area between the parabola $\mathrm{y}={\mathrm{x}}^2$and the straight line 2x - y + 8 = 0.

Intersecting points identification

From the figure, the intersecting points are:

$$\begin{gathered} x_{1,2}=\frac{2\pm \sqrt{4+32}}{2} \\ =(4,-2) \end{gathered}$$

Calculation of the area under the curve

The integral evaluates to:

$$\begin{gathered} \int {\int }_{\mathrm{A}}xdxdy={\int }_{x=-2}^4\left[{\int }_{y=x^2}^{2x+8}dy\right]xdx \\ ={\int }_{-2}^4\left(2x+8-x^2\right)dx \\ ={\overline{)\left(\frac{2}{3}{x}^3+4x^2-\frac{1}{4}{x}^4\right)}}_{-2}^4 \\ =36 \end{gathered}$$

Therefore, the area under the given graph is 36.