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

Find Derivative of H(x): First, find the derivative of H(x) = 10^x, which is h(x). h(x) = H'(x) = ln(10) * 10^xIntegrate h(x) from -3 to 2: Now, integrate h(x) from -3 to 2. ∫(-3 to 2) ln(10) * 10^x dxPull Out Constant: Pull out the constant ln(10) from the integral. ln(10) * ∫(-3 to 2) 10^x dxIntegrate 10^x: Integrate 10^x with respect to x. ∫10^x dx = (10^x) / ln(10) + CEvaluate Integral: Now, evaluate the integral from -3 to 2. ln(10) * [(10^2) / ln(10) - (10^(-3)) / ln(10)]Simplify Expression: Simplify the expression. ln(10) * [100/ln(10) - 1/(1000ln(10))]Cancel out ln(10): Cancel out the ln(10) in the numerator and denominator. 100 - 1/1000Combine Terms: Combine the terms to get the final answer. 100 - 0.001 = 99.999

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H(x)=10^(x)

h(x)=H^(')(x)

int_(-3)^(2)h(x)dx=

$H(x)=10^{x}$ $\newline$ $h(x)=H^{\prime}(x)$ $\newline$ $\int_{-3}^{2} h(x) d x=$

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Calculus

Evaluate definite integrals using the power rule

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H(x)=10^{x}

\newline

h(x)=H^{\prime}(x)

\newline

\int_{-3}^{2} h(x) d x=

Find Derivative of H(x): First, find the derivative of $H(x) = 10^x$ , which is $h(x)$ . $h(x) = H'(x) = \ln(10) \cdot 10^x$
Integrate $h(x)$ from $-3$ to $2$ : Now, integrate $h(x)$ from $-3$ to $2$ . $\newline$ $\int_{-3}^{2} \ln(10) \cdot 10^x \, dx$
Pull Out Constant: Pull out the constant $\ln(10)$ from the integral. $\newline$ $\ln(10) \times \int_{-3}^{2} 10^x \, dx$
Integrate $10^x$ : Integrate $10^x$ with respect to $x$ . $\int 10^x dx = \frac{10^x}{\ln(10)} + C$
Evaluate Integral: Now, evaluate the integral from $-3$ to $2$ . $\ln(10) \times \left[\frac{10^2}{\ln(10)} - \frac{10^{-3}}{\ln(10)}\right]$
Simplify Expression: Simplify the expression. $\ln(10) \times \left[\frac{100}{\ln(10)} - \frac{1}{1000\ln(10)}\right]$
Cancel out $\ln(10)$ : Cancel out the $\ln(10)$ in the numerator and denominator. $\newline$ $100 - \frac{1}{1000}$
Combine Terms: Combine the terms to get the final answer. $100 - 0.001 = 99.999$