Explanation
=(−2x33+x2)|10−∫xlogxdxto Integrate xlogxdx, substitute logx=t, dx=xdtIntegral becomes ∫e2ttdtIntegration by parts and then putting limits gives =−1/4
The ratio in which the area bounded by the curves y^2=12x and x^2=12y is divided by the line x = 3 is
y^2 =12 x x^2 =12 y \displaystyle \int_0^3 \sqrt{12 x}-\int_0^3 \dfrac{x^2}{12} =\int_0^3 \sqrt{12 x} -\dfrac{x^2}{12} \displaystyle \dfrac{2\sqrt{12}x^{\dfrac{3}{2}}}{3} - \dfrac{x^3}{36}\int_0^3 \displaystyle \dfrac{\sqrt{12}3\sqrt{3}}{3} - \dfrac{27}{36} 12 - \dfrac{3}{4} =\dfrac{45}{4} Same as \dfrac{8}{4} for remaining part \therefore ratio =\dfrac{15}{49}
Locus of |z-(4+4i)|=4 is a circle with center at (4,4) and radius 4 is Complex plane.
Hence, locus of |z-(4+4i)| \geq 4 is all points either on or outside the circle with radius 4 and center (4,4).
Similarly, locus of |-z-(4+4i)| \geq 4 is all points on or outside the circle with radius 4 and center (-4,-4).
Locus of |iz-(4+4i)| \geq 4 is all points on or outside the circle with radius 4 and center (-4,4).
Finally, locus of |-iz-(4+4i)| \geq 4 is all points on or outside the circle with radius 4 and center (4,-4).
Hence, the area bounded by locus of all four will be the area enclosed by the four circles in argand plane as shown in the figure. Area bounded= area of shaded region= 64 - \pi r^{2}=64-16\pi =16(4-\pi )
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