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QJEEAdv06Mac587: JEE Advanced 2006 Mathematics - Class 12 JEEAdv06Mac587 Solution: Let the definite integral be defined by the formula For more…

Let the definite integral be defined by the formula
∫abf(x)dx=b−a2(f(a)+f(b)).\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 2}\left( {f\left( a \right) + f\left( b \right)} \right).} For more accurate result for
c∈(a,b),c \in \left( {a,b} \right), we can use ∫abf(x)dx=∫acf(x)dx+∫cbf(x)dx=F(c)\int\limits_a^b {f\left( x \right)dx = \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx = F\left( c \right)} } } so
that for c=a+b2,c = {{a + b} \over 2}, we get ∫abf(x)dx=b−a4(f(a)+f(b)+2f(c)).\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 4}\left( {f\left( a \right) + f\left( b \right) + 2f\left( c \right)} \right).}

∫0π/2sin⁡x dx=\int\limits_0^{\pi /2} {\sin x\,dx = }

Multiple Choice Options

  • A π8(1+2){\pi \over 8}\left( {1 + \sqrt 2 } \right)
  • B π4(1+2){\pi \over 4}\left( {1 + \sqrt 2 } \right)
  • C π82{\pi \over {8\sqrt 2 }}
  • D π42{\pi \over {4\sqrt 2 }}

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