What is the arclength of f(x)=2-3x f(x)=2−3x in the interval [-2,1][−2,1]?

1 Answer
Rui D.
Apr 10, 2016

3sqrt103√10

Explanation:

f(x)=2-3xf(x)=2−3x
f"'"(x)=-3f'(x)=−3

L=int_a^b sqrt(1+[f"'"(x)]^2)*dxL=∫ba√1+[f'(x)]2⋅dx
L=int_-2^1 sqrt(1+9)*dxL=∫1−2√1+9⋅dx
L=sqrt10*x|_-2^1=sqrt10(1-(-2))=sqrt10(1+2)=3sqrt10L=√10⋅x∣1−2=√10(1−(−2))=√10(1+2)=3√10

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