A admins Administrator Staff member Jan 29, 2022 #1 \[\text{ If A =}\left(5xy-6x^2\right)i+(2y-4x)j\]Evaluate \( \int_{C} A \cdot d x \) where \(C\) is the curve \( y=x^{3} \) in any plane from the point \( (1,1) \) to \( (2,8) \). Last edited by a moderator: Jan 31, 2025
\[\text{ If A =}\left(5xy-6x^2\right)i+(2y-4x)j\]Evaluate \( \int_{C} A \cdot d x \) where \(C\) is the curve \( y=x^{3} \) in any plane from the point \( (1,1) \) to \( (2,8) \).
S solution.livedoubt New member Staff member Mar 1, 2025 #2 \(y=x^3\) \(dy=3x^2dx\) \[\begin{aligned}\int_{c}A\cdot dx & =\int_1^2\left(5xy-6x^2\right)dx+(2y-4x)dy\\ & =\int_1^2\left(5xx^3-6x^2\right)dx+\left(2x^3-4x\right)\left(3x^2dx\right)\\ & =\int_1^2\left(5x^4-6x^2+6x^5-12x^3\right)dx\\ & =\left[\frac{5x^5}{5}-\frac{6 x^{3}}{3}+\frac{6 x^{6}}{6}-\frac{12 x^{4}}{4}\right]_1^2\\ & =\left[x^5-2x^3+x^6-3x^4\right]_1^2\\ & =[32-2(8)+64-3(16)]-[1-2+1-3]\\ & =[32]-[-3]\\ & =32+3=35\end{aligned}\] Therefore, \[\int_{c}A\cdot dx=35\]
\(y=x^3\) \(dy=3x^2dx\) \[\begin{aligned}\int_{c}A\cdot dx & =\int_1^2\left(5xy-6x^2\right)dx+(2y-4x)dy\\ & =\int_1^2\left(5xx^3-6x^2\right)dx+\left(2x^3-4x\right)\left(3x^2dx\right)\\ & =\int_1^2\left(5x^4-6x^2+6x^5-12x^3\right)dx\\ & =\left[\frac{5x^5}{5}-\frac{6 x^{3}}{3}+\frac{6 x^{6}}{6}-\frac{12 x^{4}}{4}\right]_1^2\\ & =\left[x^5-2x^3+x^6-3x^4\right]_1^2\\ & =[32-2(8)+64-3(16)]-[1-2+1-3]\\ & =[32]-[-3]\\ & =32+3=35\end{aligned}\] Therefore, \[\int_{c}A\cdot dx=35\]
