Answer to Question #345159 in Calculus for john

Question #345159

find the area of the region in the first quadrant by bounded by the curve y=sqrt x^2+5, the x-axis and the line x=2


1
Expert's answer
2022-05-27T00:26:20-0400

ANSWER: Area of region 5.0118\cong 5.0118

EXPLANATION:

Let D={(x,y):0x2,0yx2+5},A=AreaD=\left \{ (x,y):0\leq x\leq2,\, 0\leq y\leq \sqrt{x^{2}+5} \right \} , A=Area of DD , then

A=02x2+5dxA=\int_{0}^{2}\sqrt{x^{2}+5}dx

We calculate the integral by parts : denote u=x2+5,dv=dxu= \sqrt{x^{2}+5} ,\, dv=dx\, . Therefore , v=x,du=2x2x2+5=xx2+5v=x, \, du=\frac{2x}{2\sqrt{x^{2}+5}}=\frac{ x}{ \sqrt{x^{2}+5}} and

A=[xx2+5]0202x2x2+5dx=[222+50]02x2+55x2+5dx=6A+5021x2+5dx=6A+5[lnx+x2+5]02=6A+5(ln(2+3)ln5)=6A+5ln55=6A+512ln5.A=\left [ x\cdot \sqrt{x^{2}+5} \right ]_{0}^{2}-\int_{0}^{2}\frac{x^{2}}{\sqrt{x^{2}+5}}dx=\left [ 2\cdot \sqrt{2^{2}+5}-0 \right ] -\int_{0}^{2}\frac{x^{2}+5-5}{\sqrt{x^{2}+5}}dx=6-A+5\cdot \int_{0}^{2}\frac{1}{\sqrt{x^{2}+5}}dx=6-A+5\cdot \left [ \ln \left | x+\sqrt{x^{2}+5} \right | \right ]_{0}^{2}=6-A+5\cdot \left ( \ln (2+3) -\ln \sqrt{5}\right )=6-A+5\ln \frac{5}{\sqrt{5}}=6-A+5\cdot\frac{1}{2}\ln 5 .

From the equality A=6A+512ln5A=6-A+5\cdot\frac{1}{2}\ln 5 it follows A=3+54ln55.0118.A=3+\frac{5}{4}\ln 5 \cong 5.0118.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment