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Using the Fundamental Theorem for line
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Calculus: Early Transcendentals (3rd Edition)
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- 5. Prove that the equation has no solution in an ordered integral domain.arrow_forwardAssume that F= (e^x−y)i+ (2y−x)j, C1 is curve from (0,1) to (2,5) along the graph of y=x^2+ 1, and C2 is the line segment from (0,1) to (2,5).(a) Parameterize C2and compute ∫C2F·dr. Must it be the case that the value you obtained for∫C2F·dr is equaled to that of ∫C1F·dr? Explain. b) Use the Fundamental Theorem of Line Integrals to compute ∫C1F·dr or explain why this can not be done.arrow_forwarda) Use the Trapezoid Rule to estimate the following integral. Use n=5 integral of (e^(x^2)) dx from 0 to 1 b) Use the Error Bound Formula for the Trapezoid Rule to find a best estimate for the error involved in part aarrow_forward
- Use Green's Theorem to evaluate the integral of F · dr. over C (Check the orientation of the curve before applying the theorem.) F(x, y) = (y2 cos(x), x2 + 2y sin(x)) C is the triangle from (0, 0) to (1, 3) to (1, 0) to (0, 0)arrow_forwardGeneral construction of Riemann integral of function f in ℝ³arrow_forwardUsing integral, where , convert the line integral ∮ 2 to a double Green’s theorem C is the boundary of the square with vertices (2, 2) and (2, -2). ( do not evaluate the integral)arrow_forward
- Green’s Theorem for line integrals Use either form of Green’sTheorem to evaluate the following line integral.arrow_forwardUsing the Fundamental Theorem for line integrals Verifythat the Fundamental Theorem for line integral can be used to evaluatethe given integral, and then evaluate the integral.arrow_forwardDetermine the necessary or sufficient conditions for the ∫P(x)/q(x) dxBe an immediate integral, where p(x) is a grade 2 polynomial and q(x) a grade 3 polynomial Considerations1. Let g be a Real root of q(x). Then q(x)=(x-g) q1(x), where q1(x) is a polynomial of degree 2. 2. Find constants s,y,m so that p(x)=ax2+bx+c = s(q'(x)) + y(x-g) + m 3. Without loss of generality we can consider q(x)= fx3 +hx2 +jx +k In the end, this means that∫ p(x)/q(x)dx=s∫ q'(x)/q(x)dx+y∫(x-g)/q(x)dx+∫ (m)/q(x) = s∫q'(x)/q(x)dx+y∫dx/q1(x) + m∫dx/q(x). Depending on whether q1(x) is an irreducible quadratic polynomial or not, it can be seen that the above can easily be expressed as s ln (q(x)) + y ∫ dx/q1(x) + m∫dx/q(x) Note: Don't use the partial fraction method to integrate and a numerical examplearrow_forward
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- Elements Of Modern AlgebraAlgebraISBN:9781285463230Author:Gilbert, Linda, JimmiePublisher:Cengage Learning,