The area bounded by the curves y=frac{1}{4}le | vedclass

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(B) The given curves are $y=\frac{1}{4}|4-x^2|$ and $y=7-|x|$. Both curves are symmetric about the $y$-axis.The area of the shaded region is given by

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) The given curves are $y=\frac{1}{4}|4-x^2|$ and $y=7-|x|$. Both curves are symmetric about the $y$-axis.The area of the shaded region is given by $A = 2 \int_{0}^{4} (y_{upper} - y_{lower}) dx$.For $x \in [0, 4]$,$y_{upper} = 7-x$ and $y_{lower} = \frac{1}{4}|4-x^2|$.Thus,$A = 2 \int_{0}^{4} (7-x - \frac{1}{4}|4-x^2|) dx = 2 \left[ \int_{0}^{4} (7-x) dx - \frac{1}{4} \int_{0}^{4} |4-x^2| dx \right]$.Calculating the first integral: $\int_{0}^{4} (7-x) dx = [7x - \frac{x^2}{2}]_{0}^{4} = 28 - 8 = 20$.Calculating the second integral: $\int_{0}^{4} |4-x^2| dx = \int_{0}^{2} (4-x^2) dx + \int_{2}^{4} (x^2-4) dx$.$= [4x - \frac{x^3}{3}]_{0}^{2} + [\frac{x^3}{3} - 4x]_{2}^{4} = (8 - \frac{8}{3}) + ((\frac{64}{3} - 16) - (\frac{8}{3} - 8)) = \frac{16}{3} + (\frac{16}{3} - (-\frac{16}{3})) = \frac{16}{3} + \frac{32}{3} = \frac{48}{3} = 16$.Substituting these values back into the area formula:$A = 2 [20 - \frac{1}{4}(16)] = 2 [20 - 4] = 2 [16] = 32$.Therefore,the area is $32$ square units.

The area bounded by the curves $y=\frac{1}{4}\left|4-x^2\right|$ and $y=7-|x|$ is

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