If the normal to the curve y(x)=int_{0}^{x}(2 | vedclass

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(B) Given the curve $y(x)=\int_{0}^{x}(2t^{2}-15t+10)dt$.By the Fundamental Theorem of Calculus,the slope of the tangent at $x=a$ is $y'(a) = 2a^{2}-1

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$406$

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(B) Given the curve $y(x)=\int_{0}^{x}(2t^{2}-15t+10)dt$.By the Fundamental Theorem of Calculus,the slope of the tangent at $x=a$ is $y'(a) = 2a^{2}-15a+10$.The slope of the normal at $(a, b)$ is $-\frac{1}{y'(a)} = -\frac{1}{2a^{2}-15a+10}$.The given line is $x+3y=-5$,which can be written as $y=-\frac{1}{3}x-\frac{5}{3}$. The slope of this line is $-\frac{1}{3}$.Since the normal is parallel to the line,their slopes are equal:$-\frac{1}{2a^{2}-15a+10} = -\frac{1}{3} \Rightarrow 2a^{2}-15a+10 = 3$.$2a^{2}-15a+7 = 0 \Rightarrow (2a-1)(a-7) = 0$.Since $a>1$,we have $a=7$.Now,calculate $b = y(7) = \int_{0}^{7}(2t^{2}-15t+10)dt = [\frac{2}{3}t^{3} - \frac{15}{2}t^{2} + 10t]_{0}^{7}$.$b = \frac{2}{3}(343) - \frac{15}{2}(49) + 10(7) = \frac{686}{3} - \frac{735}{2} + 70 = \frac{1372 - 2205 + 420}{6} = -\frac{413}{6}$.Thus,$6b = -413$.$|a+6b| = |7 - 413| = |-406| = 406$.

If the normal to the curve $y(x)=\int_{0}^{x}(2t^{2}-15t+10)dt$ at a point $(a, b)$ is parallel to the line $x+3y=-5$,where $a>1$,then the value of $|a+6b|$ is equal to..........

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