The area of the region bounded by the curve y | vedclass

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(A) The required area is given by the definite integral of the function $y$ with respect to $x$ from $x=1$ to $x=5$. Since the curve is above the $x$-

Vedclass · Accepted Answer

$428 	ext{ sq. units}$

Vedclass · Answer

(A) The required area is given by the definite integral of the function $y$ with respect to $x$ from $x=1$ to $x=5$. Since the curve is above the $x$-axis in the interval $[1, 5]$,the area is: $	ext{Area} = \int_{1}^{5} (4x^{3}-6x^{2}+4x+1) dx$ $= \left[\frac{4x^{4}}{4} - \frac{6x^{3}}{3} + \frac{4x^{2}}{2} + xight]_{1}^{5}$ $= \left[x^{4} - 2x^{3} + 2x^{2} + xight]_{1}^{5}$ $= [5^{4} - 2(5)^{3} + 2(5)^{2} + 5] - [1^{4} - 2(1)^{3} + 2(1)^{2} + 1]$ $= [625 - 250 + 50 + 5] - [1 - 2 + 2 + 1]$ $= 430 - 2 = 428 	ext{ sq. units}$ Thus,the correct option is $A$.

The area of the region bounded by the curve $y=4x^{3}-6x^{2}+4x+1$ and the lines $x=1, x=5$ and the $x$-axis is

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