Answer in Calculus for zakkir sharma #204868

Evaluate _C∫ (2x² + 3y²)dx, where C is the curve given by x(t)=at² , y(t)=2at, 0≤t≤1.

Let us evaluate “int_C (2x^2 + 3y^2)dx”, where “C” is the curve given by “x(t)=at^2 , y(t)=2at, 0u2264tu22641.”

“int_C (2x^2 + 3y^2)dx=int_0^1 (2(at^2)^2 + 3(2at)^2)d(at^2)=int_0^1 (2a^2t^4 + 12a^2t^2)2atdt=nint_0^1 (4a^3t^5 + 24a^3t^3)dt=4a^3int_0^1 (t^5 + 6t^3)dt=4a^3(frac{t^6}{6}+6frac{t^4}{4})|_0^1=n4a^3(frac{1}{6}+frac{3}{2})=4a^3frac{10}{6}=frac{20a^3}{3}.”