To evaluate the double integral \[\int_0^8 {\left( {\int_{\frac{{\text{y}}}{2}}^{\frac{{\text{y}}}{2} + 1} {\left( {\frac{{2{\text{x}} - {\text{y}}}}{2}} \right){\text{dx}}} } \right){\text{dy,}}} \] we make the substitution \[{\text{u}} = \frac{{2{\text{x}} - {\text{y}}}}{2}\] and \[{\text{v}} = \frac{{\text{y}}}{2}.\] The integral will reduce to
A. \[\int_0^4 {\left( {\int_0^2 {2{\text{u du}}} } \right){\text{dv}}} \]
B. \[\int_0^4 {\left( {\int_0^1 {2{\text{u du}}} } \right){\text{dv}}} \]
C. \[\int_0^4 {\left( {\int_0^1 {{\text{u du}}} } \right){\text{dv}}} \]
D. \[\int_0^4 {\left( {\int_0^2 {{\text{u du}}} } \right){\text{dv}}} \]
Answer: Option B
This Question Belongs to Engineering Maths >> Calculus
The Taylor series expansion of 3 sinx + 2 cosx is . . . . . . . .
A. 2 + 3x - x2 - \[\frac{{{{\text{x}}^3}}}{2}\] + ...
B. 2 - 3x + x2 - \[\frac{{{{\text{x}}^3}}}{2}\] + ...
C. 2 + 3x + x2 + \[\frac{{{{\text{x}}^3}}}{2}\] + ...
D. 2 - 3x - x2 + \[\frac{{{{\text{x}}^3}}}{2}\] + ...
B. \[\infty \]
C. \[\frac{1}{2}\]
D. \[ - \infty \]
A. \[1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
B. \[ - 1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
C. \[1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
D. \[ - 1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
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