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in Definite Integrals by (15 points)
edited by

If F = xyi + (x+ y2) j evaluate f. dr along the curve c y = x2 - 4 in the xy plane from the point (2, 0) (4, 12).

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1 Answer

+1 vote
by (49.0k points)

\(\vec F = xy\hat i + (x^2 + y^2)\hat j\)

\(\vec F.\vec {dr} = (xy\hat i + (x^2 + y^2)\hat j).(dx\hat i +dy\hat j)\)

\(= xydx + (x^2 + y^2)dy\)

Required

\(= \int \limits_c\vec F.\vec {dr}\), where \(c:y = x^2 - 4\)

\(= \int\limits_2^4x(x^2 -4)dx + (x^2 + (x^2 - 4)^2)(2xdx)\)

\(= \int\limits_2^4(x^3 -4x)dx + (2x^5 - 14x^3 + 32x)dx\)

\(= \int\limits_2^4 (2x^5 - 13x^3 + 28x)dx\)

\(= \frac 26 [x^6]_2^4 - \frac{13}4 [x^4]_2^4 + \frac{28}2[x^2]_2^4\)

\(= \frac 13 (4096 - 64) - \frac{13}4 (256 - 16 ) + 14(16 - 4)\)

\(= 1344 -780 + 168\)

\(= 1512 - 780\)

\(= 732\)

by (10 points)
Why there is 2xdx...
by (45.5k points)
The integration of 2x in calculus is equal to x square plus the constant of integration which is symbolically written as ∫2x dx = x^2 + C, where ∫ is the symbol of the integral, dx shows that the integration of 2x is with respect to the variable x and C is the constant of integration.
by (10 points)
2x^5-14x^3+32x   where it came from

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