Changing the order of the integration in the double integral \(I
Changing the order of the integration in the double integral \(I = \mathop \smallint \limits_0^8 \mathop \smallint \limits_{\frac{x}{4}}^2 f\left( {x,\;y} \right)dy~dx\), leads to \(I =I = \mathop \smallint \limits_r^s \mathop \smallint \limits_p^q f\left( {x,\;y} \right)dx~dy\)
Which of the following conclusions is/are correct?A. The value of p will be 0
B. The value of q will be 16 y<sup style="">2</sup>
C. The value of r will be 2
D. The value of s will be 2
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Right Answer is:
SOLUTION
\(I = \mathop \smallint \limits_0^8 \mathop \smallint \limits_{\frac{x}{4}}^2 f\left( {x,\;y} \right)dy~dx\)
We can draw the graph from the limits of the integration.
The limit of y is from \(y = \frac{x}{4}\) to y = 2. For x, the limit is x = 0 to x = 8
Here we change the order of the integration. The limit of x is 0 to 8, but we have to find the limits in the form of y, i.e. x = 0 to x = 4y, and the limit of y is 0 to 2.
\(\mathop \smallint \nolimits_0^8 \mathop \smallint \nolimits_{\frac{x}{4}}^2 f\left( {x,\;y} \right)dydx = \mathop \smallint \nolimits_0^2 \mathop \smallint \nolimits_0^{4y} f\left( {x,\;y} \right)dxdy\)
\(= \mathop \smallint \nolimits_r^s \mathop \smallint \nolimits_p^q f\left( {x,\;y} \right)dxdy\)
Comparing the limits, we get:
r = 0, s = 2, p = 0, q = 4y
