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{:[H(x)=|(x)/(2)|],[h(x)=H^(')(x)],[int_(-6)^(-2)h(x)dx=]:}

H(x)=∣x2∣h(x)=H′(x)∫−6−2h(x)dx= \begin{array}{l}H(x)=\left|\frac{x}{2}\right| \\ h(x)=H^{\prime}(x) \\ \int_{-6}^{-2} h(x) d x=\end{array}

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Full solution

Q. H(x)=∣x2∣h(x)=H′(x)∫−6−2h(x)dx= \begin{array}{l}H(x)=\left|\frac{x}{2}\right| \\ h(x)=H^{\prime}(x) \\ \int_{-6}^{-2} h(x) d x=\end{array}
  1. Find Derivative of H(x): First, we need to find the derivative of H(x) which is h(x).\newlineH(x) = ∣x2∣\left|\frac{x}{2}\right|, so we need to consider the absolute value.
  2. Consider Absolute Value: For x < 0, H(x)=−x2H(x) = -\frac{x}{2}, so h(x)=ddx(−x2)=−12h(x) = \frac{d}{dx}(-\frac{x}{2}) = -\frac{1}{2}. For x > 0, H(x)=x2H(x) = \frac{x}{2}, so h(x)=ddx(x2)=12h(x) = \frac{d}{dx}(\frac{x}{2}) = \frac{1}{2}. Since we're integrating from −6-6 to −2-2, we only need h(x)h(x) for x < 0.
  3. Integrate h(x)h(x) from −6-6 to −2-2: Now we integrate h(x)h(x) from −6-6 to −2-2. ∫−6−2−12 dx=−12×x∣−6−2\int_{-6}^{-2} -\frac{1}{2} \, dx = -\frac{1}{2} \times x \bigg|_{-6}^{-2}.
  4. Plug in Limits: Plug in the limits of integration.\newline(−12×−2)−(−12×−6)=1−3=−2(-\frac{1}{2} \times -2) - (-\frac{1}{2} \times -6) = 1 - 3 = -2.

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