To evaluate the double integral $\mathop \smallint \limits_0^8 \left( {\mathop \smallint \limits_{\frac{y}{2}}^{\left( {\frac{y}{2}} \right) + 1} \left( {\frac{{2x - y}}{2}} \right)dx} \right)dy$ we make the substitution $u = \left( {\frac{{2x - y}}{2}} \right)$ and $v = \frac{y}{2}$. The integral will reduce to

Question

To evaluate the double integral $\mathop \smallint \limits_0^8 \left( {\mathop \smallint \limits_{\frac{y}{2}}^{\left( {\frac{y}{2}} \right) + 1} \left( {\frac{{2x - y}}{2}} \right)dx} \right)dy$ we make the substitution $u = \left( {\frac{{2x - y}}{2}} \right)$ and $v = \frac{y}{2}$. The integral will reduce to

TuteeHUB · Accepted Answer

$\mathop \smallint \limits_0^4 \left( {\mathop \smallint \limits_0^1 \;u\;du} \right)dv$

TuteeHUB · Answer

$\mathop \smallint \limits_0^4 \left( {\mathop \smallint \limits_0^2 2\;u\;du} \right)dv$

TuteeHUB · Answer

$\mathop \smallint \limits_0^4 \left( {\mathop \smallint \limits_0^1 2\;u\;du} \right)dv$

TuteeHUB · Answer

$\mathop \smallint \limits_0^4 \left( {\mathop \smallint \limits_0^2 \;u\;du} \right)dv$

1.	To evaluate the double integral \(\mathop \smallint \limits_0^8 \left( {\mathop \smallint \limits_{\frac{y}{2}}^{\left( {\frac{y}{2}} \right) + 1} \left( {\frac{{2x - y}}{2}} \right)dx} \right)dy\) we make the substitution \(u = \left( {\frac{{2x - y}}{2}} \right)\) and \(v = \frac{y}{2}\). The integral will reduce to
A.	\(\mathop \smallint \limits_0^4 \left( {\mathop \smallint \limits_0^2 2\;u\;du} \right)dv\)
B.	\(\mathop \smallint \limits_0^4 \left( {\mathop \smallint \limits_0^1 2\;u\;du} \right)dv\)
C.	\(\mathop \smallint \limits_0^4 \left( {\mathop \smallint \limits_0^1 \;u\;du} \right)dv\)
D.	\(\mathop \smallint \limits_0^4 \left( {\mathop \smallint \limits_0^2 \;u\;du} \right)dv\)
Answer» C. \(\mathop \smallint \limits_0^4 \left( {\mathop \smallint \limits_0^1 \;u\;du} \right)dv\)

Discussion

No Comment Found

Related MCQs

Reply to Comment