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$Hitung integral lipat dua berikut ini: {:[∬_(R)(2x+3y)dA],[" dimana "R={(x","y)∣1 <= x <= 2","0 <= y <= 3}]:}$

$1$ . Hitung integral lipat dua berikut ini: $\newline$ $\begin{array}{l} \iint_{R}(2 x+3 y) d A \\ \text { dimana } R=\{(x, y) \mid 1 \leq x \leq 2,0 \leq y \leq 3\} \end{array}$

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Roots of rational numbers

Full solution

Q.

1

. Hitung integral lipat dua berikut ini:

\newline

\begin{array}{l} \iint_{R}(2 x+3 y) d A \\ \text { dimana } R=\{(x, y) \mid 1 \leq x \leq 2,0 \leq y \leq 3\} \end{array}

Identify limits of integration: Identify the limits of integration for the region $R$ . For $x$ , the limits are from $1$ to $2$ . For $y$ , the limits are from $0$ to $3$ .
Set up double integral: Set up the double integral. $\iint_{R}(2x+3y)\,dA = \int_{1}^{2}\int_{0}^{3}(2x+3y)\,dy\,dx$
Integrate with respect to y: Integrate with respect to y first. $\newline$ $\int_{0}^{3}(2x+3y)dy = 2x\int_{0}^{3}dy + 3\int_{0}^{3}y dy$ $\newline$ $= 2x[y]_{0}^{3} + 3\left[\frac{y^2}{2}\right]_{0}^{3}$ $\newline$ $= 2x(3) + 3\left(\frac{3^2}{2}\right) - 3\left(\frac{0^2}{2}\right)$ $\newline$ $= 6x + \frac{27}{2}$
Integrate with respect to x: Now integrate with respect to x. $\newline$ $\int_{1}^{2}(6x + \frac{27}{2})dx = 6\int_{1}^{2}xdx + (\frac{27}{2})\int_{1}^{2}dx$ $\newline$ = $6[x^2/2]_{1}^{2} + (\frac{27}{2})[x]_{1}^{2}$ $\newline$ = $6(2^2/2 - 1^2/2) + (\frac{27}{2})(2 - 1)$ $\newline$ = $6(2) + \frac{27}{2}$ $\newline$ = $12 + \frac{27}{2}$ $\newline$ = $12 + 13.5$ $\newline$ = $25.5$

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