Let I-c-Rxy2dxdy- where R is the region shown in the figure and c-6- 10-4 - The value of I equals Given the answer up to two decimal places-

I=c∫∫_Rxy^2dxdy Region R is bounded by y=0 and y=2x I=c [ ∫_1^5∫_0^2xxy^2dy dx ] =c [ ∫_1^5. ( xy^33 ) |_0^2xdx ]=c [ ∫_1^583x^4dx ] =c [. nbsp;815x^5 |_ ...

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Let I=c∫∫Rxy2...

Question

Let $I = c \int \int_{R} x y^{2} d x d y,$ where $R$ is the region shown in the figure and $c = 6 \times 10^{- 4}$ . The value of $I$ equals (Given the answer up to two decimal places).

0.99

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Solution

The correct option is A 0.99
$I = c \int \int_{R} x y^{2} d x d y$

Region $R$ is bounded by
$y = 0$ and $y = 2 x$
$I = c [\int_{1}^{5} \int_{0}^{2 x} x y^{2} d y d x]$
$= c ⎡ ⎣ \int_{1}^{5} {\begin{matrix} (\frac{x y^{3}}{3}) \end{matrix} ∣ ∣ ∣ ∣}_{0}^{2 x} d x ⎤ ⎦ = c [\int_{1}^{5} \frac{8}{3} x^{4} d x]$
$= c [{\begin{matrix} \frac{8}{15} x^{5} \end{matrix} ∣ ∣ ∣}_{1}^{5}] = c [\frac{8}{15} (5^{5} - 1)]$
$= 6 \times 10^{- 4} \times [\frac{8}{15} (5^{5} - 1)]$
$I = 0.99$ (upto two decimal places)

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Let I=c∫∫Rxy2dxdy, where R is the region shown in the figure and c=6×10−4 . The value of I equals (Given the answer up to two decimal places). 0.99

The correct option is A 0.99I=c∫∫Rxy2dxdy Region R is bounded by y=0 and y=2x I=c[∫51∫2x0xy2dydx] =c⎡⎣∫51(xy33)∣∣ ∣∣2x0dx⎤⎦=c[∫5183x4dx] =c[815x5∣∣∣51]=c[815(55−1)] =6×10−4×[815(55−1)] I=0.99 (upto two decimal places)

Let $I = c \int \int_{R} x y^{2} d x d y,$ where $R$ is the region shown in the figure and $c = 6 \times 10^{- 4}$ . The value of $I$ equals (Given the answer up to two decimal places).

0.99