If int limits_{-0.15}^{0.15} |100 x^2 - 1| dx | vedclass

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(D) Since the integrand $f(x) = |100x^2 - 1|$ is an even function,we have $\int_{-0.15}^{0.15} |100x^2 - 1| dx = 2 \int_{0}^{0.15} |100x^2 - 1| dx$.Th

Vedclass · Accepted Answer

$575$

Vedclass · Answer

(D) Since the integrand $f(x) = |100x^2 - 1|$ is an even function,we have $\int_{-0.15}^{0.15} |100x^2 - 1| dx = 2 \int_{0}^{0.15} |100x^2 - 1| dx$.The critical point is $100x^2 - 1 = 0$,which gives $x^2 = \frac{1}{100}$,so $x = 0.1$ (within the interval $[0, 0.15]$).Thus,$I = 2 \left[ \int_{0}^{0.1} (1 - 100x^2) dx + \int_{0.1}^{0.15} (100x^2 - 1) dx \right]$.Evaluating the integrals:$I = 2 \left[ x - \frac{100x^3}{3} \right]_0^{0.1} + 2 \left[ \frac{100x^3}{3} - x \right]_{0.1}^{0.15}$.$I = 2 \left( 0.1 - \frac{100(0.001)}{3} \right) + 2 \left( (\frac{100(0.003375)}{3} - 0.15) - (\frac{100(0.001)}{3} - 0.1) \right)$.$I = 2 \left( 0.1 - \frac{0.1}{3} \right) + 2 \left( 0.1125 - 0.15 - \frac{0.1}{3} + 0.1 \right)$.$I = 2 \left( \frac{0.2}{3} \right) + 2 \left( 0.0625 - \frac{0.1}{3} \right) = \frac{0.4}{3} + 0.125 - \frac{0.2}{3} = \frac{0.2}{3} + 0.125$.$I = \frac{0.2 + 0.375}{3} = \frac{0.575}{3} = \frac{575}{3000}$.Comparing with $\frac{k}{3000}$,we get $k = 575$.

If $\int \limits_{-0.15}^{0.15} |100 x^2 - 1| dx = \frac{k}{3000}$,then $k$ is equal to $..........$.

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