find the area of the region under y = 7ln(4x) and above y = 3 for 4 ≤ x ≤ 7. round your answer to three…

Explanation:

Step1: Set up the integral

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}(f(x)-g(x))dx$. Here, $f(x)=7\ln(4x)$ and $g(x) = 3$, $a = 4$, $b=7$. So $A=\int_{4}^{7}(7\ln(4x)-3)dx$.

Step2: Split the integral

Using the property $\int_{a}^{b}(u - v)dx=\int_{a}^{b}u dx-\int_{a}^{b}v dx$, we have $A = 7\int_{4}^{7}\ln(4x)dx-3\int_{4}^{7}dx$.

Step3: Integrate $\int\ln(4x)dx$

Use integration - by - parts. Let $u=\ln(4x)$ and $dv=dx$. Then $du=\frac{1}{x}dx$ and $v = x$. By the integration - by - parts formula $\int u dv=uv-\int v du$, we get $\int\ln(4x)dx=x\ln(4x)-\int x\cdot\frac{1}{x}dx=x\ln(4x)-x + C$.

Step4: Evaluate $7\int_{4}^{7}\ln(4x)dx$

$7\int_{4}^{7}\ln(4x)dx=7\left[x\ln(4x)-x\right]_{4}^{7}=7\left[(7\ln(28)-7)-(4\ln(16)-4)\right]=7\left(7\ln(28)-7 - 4\ln(16)+4\right)=7\left(7\ln(28)-4\ln(16)-3\right)$.

Step5: Evaluate $3\int_{4}^{7}dx$

$3\int_{4}^{7}dx=3[x]_{4}^{7}=3(7 - 4)=9$.

Step6: Calculate the area

$A = 7\left(7\ln(28)-4\ln(16)-3\right)-9$. $7\ln(28)=7(\ln(4)+\ln(7))=7(2\ln(2)+\ln(7))$, $4\ln(16)=4\times4\ln(2)=16\ln(2)$. $A = 7\left(14\ln(2)+7\ln(7)-16\ln(2)-3\right)-9=7\left(- 2\ln(2)+7\ln(7)-3\right)-9$. $A=-14\ln(2)+49\ln(7)-21 - 9=-14\ln(2)+49\ln(7)-30$. Using $\ln(2)\approx0.6931$ and $\ln(7)\approx1.9459$, we get $A\approx-14\times0.6931 + 49\times1.9459-30\approx - 9.7034+95.3491 - 30\approx40.437$.

Answer:

$40.437$