Suppose we are told that $f(x)$ is a mystery function where $f((-50))=\ln (7)$ and $\lim _{x \rightarrow \infty} f(x)=0$.
Determine $\int_{-50}^{\infty} 27 f^{\prime}(x) e^{3 f(x)} d x$
The integral converges to
Answer
\(\boxed{-9*\ln(7)*7^3}\)
Step 1 ：Given that $f(x)$ is a mystery function where $f((-50))=\ln (7)$ and $\lim _{x \rightarrow \infty} f(x)=0$.
Step 2 ：We are asked to determine $\int_{-50}^{\infty} 27 f^{\prime}(x) e^{3 f(x)} d x$.
Step 3 ：This integral is in the form of $\int_{a}^{b} u dv$, where $u = 27f(x)$ and $dv = e^{3f(x)}f'(x)dx$.
Step 4 ：We can use integration by parts, which states that $\int_{a}^{b} u dv = [uv]_{a}^{b} - \int_{a}^{b} v du$. Here, $v = \frac{1}{3}e^{3f(x)}$ and $du = 27f'(x)dx$.
Step 5 ：Substituting these values into the integration by parts formula, we get $-9*f(x)*exp(3*f(-50)) + 9*f(x)*exp(3*f(\infty)) - 9*\int_{-50}^{\infty} exp(3*f(x))*f'(x) dx$.
Step 6 ：Given that $\lim _{x \rightarrow \infty} f(x)=0$ and $f(-50) = \ln(7)$, the term $9*f(x)*exp(3*f(\infty))$ will be zero because $f(\infty) = 0$.
Step 7 ：The term $-9*f(x)*exp(3*f(-50))$ will be $-9*\ln(7)*exp(3*\ln(7)) = -9*\ln(7)*7^3$.
Step 8 ：The integral term $-9*\int_{-50}^{\infty} exp(3*f(x))*f'(x) dx$ will be zero because the integrand is the derivative of $exp(3*f(x))$, which will be zero at both $-50$ and $\infty$ due to the properties of $f(x)$.
Step 9 ：So, the final answer is $-9*\ln(7)*7^3$.
Step 10 ：\(\boxed{-9*\ln(7)*7^3}\)
