The value of -03-0x6-x-ydxdy is

I=∫_0^3∫_0^x(6 x y)dxdy =∫_x=0^3∫_y=0^x(6 x y)dydx =∫_0^3(6y xy y^22)_y=0^xdx =∫_0^3(6x x^2 x^22)dx =∫_0^3(6x 3x^22)dx = ( 6x^22 x^32 )_0^3 =542 272=272=13.5 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Engineering Mathematics

Double integrals and Triple Integrals

The value of ...

Question

The value of $\int_{0}^{3} \int_{0}^{x} (6 - x - y) d x d y$ is

13.5

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

27.0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

54.0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $13.5$
$I = \int_{0}^{3} \int_{0}^{x} (6 - x - y) d x d y$
$= \int_{x = 0}^{3} \int_{y = 0}^{x} (6 - x - y) d y d x$
$= \int_{0}^{3} (6 y - x y - \frac{y^{2}}{2})_{y = 0}^{x} d x$
$= \int_{0}^{3} (6 x - x^{2} - \frac{x^{2}}{2}) d x$
$= \int_{0}^{3} (6 x - \frac{3 x^{2}}{2}) d x$
$= {(\frac{6 x^{2}}{2} - \frac{x^{3}}{2})}_{0}^{3}$
$= \frac{54}{2} - \frac{27}{2} = \frac{27}{2} = 13.5$

Suggest Corrections

Similar questions

Q. The value of integral

3 \int 0 [x] d x =

Q. The value of the integral

\int_{0}^{3} \frac{d x}{\sqrt{x + 1} + \sqrt{5 x + 1}} d x i s

\int_{0}^{\frac{π}{2}} \int_{0}^{\frac{π}{2}} s i n (x + y) d x d y

is:

Q. The double integral

\int_{0}^{a} \int_{0}^{y} f (x, y) d x d y

is equivalent to

Q. The value of

\int_{0}^{3} x^{5 / 2} d x

Join BYJU'S Learning Program

The value of ∫30∫x0(6−x−y)dxdy is

The correct option is A 13.5I=∫30∫x0(6−x−y)dxdy =∫3x=0∫xy=0(6−x−y)dydx =∫30(6y−xy−y22)xy=0dx =∫30(6x−x2−x22)dx =∫30(6x−3x22)dx =(6x22−x32)30 =542−272=272=13.5

The value of $\int_{0}^{3} \int_{0}^{x} (6 - x - y) d x d y$ is