The area bounded by the curve y=x^{3},the x-a | vedclass

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(C) The area bounded by the curve $y=f(x)$,the $x$-axis,and the lines $x=a$ and $x=b$ is given by $\int_{a}^{b} |f(x)| dx$.Here,$f(x) = x^{3}$. The cu

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$\frac{17}{4}$

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(C) The area bounded by the curve $y=f(x)$,the $x$-axis,and the lines $x=a$ and $x=b$ is given by $\int_{a}^{b} |f(x)| dx$.Here,$f(x) = x^{3}$. The curve $y=x^{3}$ crosses the $x$-axis at $x=0$.For $x \in [-2, 0]$,$x^{3} \leq 0$,and for $x \in [0, 1]$,$x^{3} \geq 0$.Therefore,the required area is $\int_{-2}^{0} |x^{3}| dx + \int_{0}^{1} |x^{3}| dx$.$= \int_{-2}^{0} (-x^{3}) dx + \int_{0}^{1} x^{3} dx$$= \left[ -\frac{x^{4}}{4} \right]_{-2}^{0} + \left[ \frac{x^{4}}{4} \right]_{0}^{1}$$= \left( 0 - \left( -\frac{(-2)^{4}}{4} \right) \right) + \left( \frac{1^{4}}{4} - 0 \right)$$= \left( 0 - (-4) \right) + \frac{1}{4}$$= 4 + \frac{1}{4} = \frac{17}{4}$ sq. units.Thus,the correct answer is $C$.

The area bounded by the curve $y=x^{3}$,the $x$-axis,and the ordinates $x=-2$ and $x=1$ is

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