h(x)=10^x\nh(x)=h(x)\n∫_{-3}^{2}h(x)dx=
h(x)=10^x\nh(x)=h(x)\n∫_{-3}^{2}h(x)dx=
Answer
Explanation:
Step1: Find the derivative of $H(x)$
The derivative of $y = a^{x}$ is $y'=a^{x}\ln a$. For $H(x)=10^{x}$, $h(x)=H'(x)=10^{x}\ln 10$.
Step2: Calculate the definite - integral
We know that $\int_{-3}^{2}h(x)dx=\int_{-3}^{2}10^{x}\ln 10dx$. Since the antiderivative of $a^{x}\ln a$ is $a^{x}$, the antiderivative of $10^{x}\ln 10$ is $10^{x}$. Then, by the fundamental theorem of calculus $\int_{-3}^{2}10^{x}\ln 10dx=10^{x}\big|_{-3}^{2}=10^{2}-10^{-3}$. $10^{2}-10^{-3}=100 - \frac{1}{1000}=\frac{100000 - 1}{1000}=\frac{99999}{1000}=99.999$.
Answer:
$99.999$