(A) The required area is given by the integral of $|y|$ with respect to $x$ from $x=-1$ to $x=2$. Since $y=x$,we have $|y|=|x|$. Required Area $= \int

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(A) The required area is given by the integral of $|y|$ with respect to $x$ from $x=-1$ to $x=2$. Since $y=x$,we have $|y|=|x|$. Required Area $= \int_{-1}^{2} |x| dx$ $= \int_{-1}^{0} |x| dx + \int_{0}^{2} |x| dx$ $= \int_{-1}^{0} (-x) dx + \int_{0}^{2} x dx$ $= \left[ -\frac{x^2}{2} \right]_{-1}^{0} + \left[ \frac{x^2}{2} \right]_{0}^{2}$ $= (0 - (-1/2)) + (4/2 - 0)$ $= 1/2 + 2 = 5/2$ sq unit.

Class 12 Mathematics Application of Integration Area bounded by region of single curve

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The area bounded by the lines $y=x$,$x=-1$,$x=2$ and the $x$-axis is

MHT CET 2008 All MHT CET

A
$5/2$ sq unit
B
$3/2$ sq unit
C
$1/2$ sq unit
D
None of the above

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Application of Integration

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Area bounded by region of single curve

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The area bounded by the lines $y=x$,$x=-1$,$x=2$ and the $x$-axis is

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