The area bounded by the curve y = -x^2,the x- | vedclass

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

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$21 	ext{ sq. units}$

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(A) The area $A$ bounded by the curve $y = f(x)$,the $x$-axis,and the lines $x = a$ and $x = b$ is given by $A = \int_{a}^{b} |f(x)| \, dx$.Here,$f(x) = -x^2$,$a = 1$,and $b = 4$.Since $y = -x^2$ lies below the $x$-axis for $x \in [1, 4]$,the area is given by:$A = \int_{1}^{4} | -x^2 | \, dx = \int_{1}^{4} x^2 \, dx$$A = \left[ \frac{x^3}{3} ight]_{1}^{4}$$A = \frac{4^3}{3} - \frac{1^3}{3}$$A = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21 	ext{ sq. units}$.

The area bounded by the curve $y = -x^2$,the $x$-axis,$x = 1$,and $x = 4$ is:

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