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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Choose the correct answer in the area bounded by the curve\( y = x^3\), the \(x\) - axis and the ordinates \(x = -2\) and \(x = 1\) is

$ \begin{array} (A) -9 \qquad &(B) \frac{-15}{4} \qquad &(C) \frac{15}{4} \qquad &(D) \frac{17}{4} \end{array} $
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1 Answer

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  • Whenever we consider areas on both the negative and positive side of the axes,we can sum them up to find the total area.
The required area is clearly on both the positive side of y-axis(i .e)in the Ist quadrant and negative side that is the IIIrd quadrant.
This is shown in the fig.
The required area is the area bounded by the curve $y=x^3$,the x-axis and the ordinate x=-2 and x=1.
Hence we can take the limits as -2 to 1.
$A=\int_{-2}^1x^3dx$
But this curve lies on both the positive side and the negative side of x-axis,let us split the limits as -2 to 0 and 0 to 1
$A=\int_{-2}^0x^3 dx+\int_0^1x^3 dx$
on integrating we get,
$A=\begin{bmatrix}\frac{x^4}{4}\end{bmatrix}_{-2}^0+\begin{bmatrix}\frac{x^4}{4}\end{bmatrix}_0^1$
on applying limits we get,
$A=\begin{bmatrix}0+\frac{(-2)^4}{4}\end{bmatrix}+\frac{1^4}{4}$ [we take $\frac{-16}{4}$ as $\frac{16}{4}$ because area cannot be negative]
$\;\;\;=\frac{16}{4}+\frac{1}{4}=\frac{17}{4}$sq.units.
Hence the required area is $\frac{17}{4}$sq.units.
The correct answer is D
answered Dec 21, 2013 by yamini.v
 

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