Question

Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,−4) to the point (1,20).

Answer 1

32.66 units

Explanation:

We are given that

`y=6x^2+14x`

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

`(dy)/(dx)=12x+14`

We know that length of curve

`s=\int_(a)^(b)\sqrt{1+((dy)/(dx))^2}dx`

We have a=-2 and b=1

Using the formula

Length of curve=
`s=\int_(-2)^(1)√(1+(12x+14)^2)dx`

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

`dt=12dx`

`dx=(1)/(12)dt`

Length of curve=
`s=(1)/(12)\int_(-2)^(1)√(1+t^2)dt`

We know that

`√(x^2+a^2)dx=\frac{x\sqrt {x^2+a^2}}{2}+(1)/(2)\ln(x+\sqrt {x^2+a^2})+C`

By using the formula

Length of curve=
`s=(1)/(12)[(t)/(2)√(1+t^2)+(1)/(2)ln(t+√(1+t^2))]^(1)_(-2)`

Length of curve=
`s=(1)/(12)[(12x+14)/(2)√(1+(12x+14)^2)+(1)/(2)ln(12x+14+√(1+(12x+14)^2))]^(1)_(-2)`

Length of curve=
`s=(1)/(12)(((12+14)√(1+(26)^2))/(2)+(1)/(2)ln(26+√(1+(26)^2))-(12(-2)+14)/(2)√(1+(-10)^2)-(1)/(2)ln(-10+√(1+(-10)^2))`

Length of curve=
`s=(1)/(12)(13√(677)+(1)/(2)ln(26+√(677))+5√(101)-(1)/(2)ln(-10+√(101))`

Length of curve=
`s=32.66`