The area under the curve y = x^2 - 4x within | vedclass

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(A) The area $A$ bounded by the curve $y = x^2 - 4x$ and the $x$-axis from $x = 0$ to $x = 2$ is given by the integral of the absolute value of the fu

Vedclass · Accepted Answer

$\frac{16}{3} 	ext{ sq. unit}$

Vedclass · Answer

(A) The area $A$ bounded by the curve $y = x^2 - 4x$ and the $x$-axis from $x = 0$ to $x = 2$ is given by the integral of the absolute value of the function. Since $x^2 - 4x \leq 0$ for $x \in [0, 2]$,the area is: $A = \left| \int_0^2 (x^2 - 4x) dx ight|$ $A = \left| \left[ \frac{x^3}{3} - 2x^2 ight]_0^2 ight|$ $A = \left| \left( \frac{8}{3} - 2(4) ight) - (0) ight|$ $A = \left| \frac{8}{3} - 8 ight| = \left| \frac{8 - 24}{3} ight| = \left| -\frac{16}{3} ight| = \frac{16}{3} 	ext{ sq. unit}$.

The area under the curve $y = x^2 - 4x$ within the $x$-axis and the line $x = 2$ is:

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