Question

evaluate the integral. \\(\int_{0}^{2} (2v + 7)(3v - 1) \\, dv\\)

Explanation:

Step1: Expand the integrand

First, we expand the product \((2v + 7)(3v - 1)\) using the distributive property (FOIL method):
\[

$$\begin{align*} (2v + 7)(3v - 1)&=2v\times3v+2v\times(-1)+7\times3v + 7\times(-1)\\ &=6v^{2}-2v + 21v-7\\ &=6v^{2}+19v - 7 \end{align*}$$

\]

Step2: Integrate term - by - term

Now we integrate the expanded function \(\int_{0}^{2}(6v^{2}+19v - 7)dv\). Using the power rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)):
\[

$$\begin{align*} \int_{0}^{2}(6v^{2}+19v - 7)dv&=\int_{0}^{2}6v^{2}dv+\int_{0}^{2}19vdv-\int_{0}^{2}7dv\\ &=6\times\frac{v^{3}}{3}\big|_{0}^{2}+19\times\frac{v^{2}}{2}\big|_{0}^{2}-7v\big|_{0}^{2}\\ &=2v^{3}\big|_{0}^{2}+\frac{19}{2}v^{2}\big|_{0}^{2}-7v\big|_{0}^{2} \end{align*}$$

\]

Step3: Evaluate the definite integral

Evaluate each term at the upper limit (\(v = 2\)) and subtract the value at the lower limit (\(v=0\)):

• For \(2v^{3}\big|_{0}^{2}\): \(2\times(2)^{3}-2\times(0)^{3}=2\times8 - 0 = 16\)
• For \(\frac{19}{2}v^{2}\big|_{0}^{2}\): \(\frac{19}{2}\times(2)^{2}-\frac{19}{2}\times(0)^{2}=\frac{19}{2}\times4-0 = 38\)
• For \(-7v\big|_{0}^{2}\): \(-7\times2-(-7\times0)=- 14 - 0=-14\)

Now sum up these results: \(16 + 38-14=40\)

Answer:

\(40\)