Answer:
32.66 units
Step-by-step explanation:
We are given that
[tex]y=6x^2+14x[/tex]
Point A=(-2,-4) and point B=(1,20)
Differentiate w.r. t x
[tex]\frac{dy}{dx}=12x+14[/tex]
We know that length of curve
[tex]s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx[/tex]
We have a=-2 and b=1
Using the formula
Length of curve=[tex]s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx[/tex]
Using substitution method
Substitute t=12x+14
Differentiate w.r t. x
[tex]dt=12dx[/tex]
[tex]dx=\frac{1}{12}dt[/tex]
Length of curve=[tex]s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt[/tex]
We know that
[tex]\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C[/tex]
By using the formula
Length of curve=[tex]s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}[/tex]
Length of curve=[tex]s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}[/tex]
Length of curve=[tex]s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})[/tex]
Length of curve=[tex]s=\frac{1}{12}(13\sqrt{677}+\frac{1}{2}ln(26+\sqrt{677})+5\sqrt{101}-\frac{1}{2}ln(-10+\sqrt{101})[/tex]
Length of curve=[tex]s=32.66[/tex]
