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H(x)=10^(x) h(x)=H^(')(x) int_(-3)^(2)h(x)dx=

H(x)=10x H(x)=10^{x} \newlineh(x)=H′(x) h(x)=H^{\prime}(x) \newline∫−32h(x)dx= \int_{-3}^{2} h(x) d x=

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Full solution

Q. H(x)=10x H(x)=10^{x} \newlineh(x)=H′(x) h(x)=H^{\prime}(x) \newline∫−32h(x)dx= \int_{-3}^{2} h(x) d x=
  1. Find Derivative of H(x): First, find the derivative of H(x)=10xH(x) = 10^x, which is h(x)h(x).h(x)=H′(x)=ln⁡(10)⋅10xh(x) = H'(x) = \ln(10) \cdot 10^x
  2. Integrate h(x)h(x) from −3-3 to 22: Now, integrate h(x)h(x) from −3-3 to 22.\newline∫−32ln⁡(10)⋅10x dx\int_{-3}^{2} \ln(10) \cdot 10^x \, dx
  3. Pull Out Constant: Pull out the constant ln⁡(10)\ln(10) from the integral.\newlineln⁡(10)×∫−3210x dx\ln(10) \times \int_{-3}^{2} 10^x \, dx
  4. Integrate 10x10^x: Integrate 10x10^x with respect to xx.∫10xdx=10xln⁡(10)+C\int 10^x dx = \frac{10^x}{\ln(10)} + C
  5. Evaluate Integral: Now, evaluate the integral from −3-3 to 22. ln⁡(10)×[102ln⁡(10)−10−3ln⁡(10)]\ln(10) \times \left[\frac{10^2}{\ln(10)} - \frac{10^{-3}}{\ln(10)}\right]
  6. Simplify Expression: Simplify the expression. ln⁡(10)×[100ln⁡(10)−11000ln⁡(10)]\ln(10) \times \left[\frac{100}{\ln(10)} - \frac{1}{1000\ln(10)}\right]
  7. Cancel out ln⁡(10)\ln(10): Cancel out the ln⁡(10)\ln(10) in the numerator and denominator.\newline100−11000100 - \frac{1}{1000}
  8. Combine Terms: Combine the terms to get the final answer. 100−0.001=99.999100 - 0.001 = 99.999

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