91Ó°ÊÓ

Use geometry to evaluate each definite integral. \(\int_{0}^{10} \frac{1}{2} x d x\)

Evaluate. $$ \int_{0}^{b} m e^{-m x} d x $$

Occasionally, integration by parts yields an integral of the form \(\int u d v\) that is identical to the original integral. In some cases, we can then solve for \(\int u d v\) algebraically. For example, to find \(\int 2^{x} e^{x} d x,\) we let \(u=2^{x}\) and \(d v=e^{x},\) so \(d u=(\ln 2) 2^{x} d x\) and \(v=e^{x} .\) Using integration by parts, we have $$ \int 2^{x} e^{x} d x=2^{x} e^{x}-\ln 2 \int 2^{x} e^{x} d x $$ Note that \(\int 2^{x} e^{x} d x\) appears twice. Adding \(\ln 2 \int 2^{x} e^{x} d x\) to $$ \begin{aligned} \int 2^{x} e^{x} d x+\ln 2 \int 2^{x} e^{x} d x &=2^{x} e^{x} \\ (1+\ln 2) \int 2^{x} e^{x} d x &=2^{x} e^{x} \\ \int 2^{x} e^{x} d x &=\frac{2^{x} e^{x}}{1+\ln 2}+C \end{aligned} $$ Use this method to evaluate the integrals in Exercises \(59-62\) $$ \begin{aligned} &\int x \ln x d x\\\ &\text { and } d v=d x \text { . Assume }\\\ &x>0 .) \end{aligned} $$

A company is producing a new product, and the time required to produce each unit decreases as workers gain experience. It is determined that $$T(x)=2+0.3\left(\frac{1}{x}\right)$$ where \(T(x)\) is the time, in hours, required to produce the \(x\) th unit. Use this information. Find the total time required for a worker to produce units 1 through 20 ; units 20 through 40 .

A car accelerates at a constant rate from 0 mph to 60 mph in 30 sec. How far has it traveled after 30 sec?

A car accelerates at a constant rate from 0 to \(60 \mathrm{mph}\) in \(30 \mathrm{sec} .\) How far does the car travel during that time?