Suppose that \( \int x \cos (3 x) d x=A x \sin (3 x)+B \cos (3 x)+C \), where \( A \) and \( B \) are constants and \( C \) is an arbitrary constant. Find the value of \( A \) and \( B \) : Answer: \( A=\square \) and \( B=\square \). Hint: A method called Integration by parts (which you will learn in the next course) can be used to find the given integral, but you do not need to use this method here. Just notice that the form of the anti- derivative is given to us. The given equation can now be expressed in a different form. Remember the definition of the indefinite integral: If \( \int f(x) d x=F(x)+C \) then \( F^{\prime}(x)=f(x) \) (i.e. \( F(x) \) is an anti-derivative of \( f(x)) \).

ón \( f_{x}=\frac{1}{2} x^{2}+x \) en el punto de abscisa \( x=a \), PASA \( a=4 \)

Encuentre \( y^{\prime} \) si \( y^{5}+3 y^{2}-2 x^{4}=-4 \)

(b) i. If \( F \) is a conservative field, prove that curl \( \vec{F}=\nabla \times \vec{F}=\overrightarrow{0} \) (i.e., \( \vec{F} \) is irrotational). ii. Conversely, if \( \nabla \times \vec{F}=\overrightarrow{0} \) (i.e., \( \vec{F} \) is irrotational), prove that \( \vec{F} \) is conservative. (c) If \( \phi=2 x y z^{2}, \quad \vec{F}=x y \vec{i}-z \vec{j}+x^{2} \vec{k} \) and C is the curve \( x=t^{2}, \quad y=2 t, \quad z=t^{3} \) from \( t=0 \) to \( t=1 \), evaluate the line integrals: i. \( \int_{C} \phi d \vec{r} \). ii. \( \int_{C} \vec{F} \times d \vec{r} \). "Mathematics Can Smile!" ( 2 mar

\( y ^ { \prime \prime \prime } + y = e ^ { x } + e ^ { - x } \cos ( 3 x ) + x ^ { 3 } \)

(f) Find the constants \( a, b, c \), so that \[ \vec{F}=(x+2 y+a z) \vec{i}+(b x-3 y-z) \vec{j}+(4 x+c y+2 z) \vec{k} \] is irrotational and hence find function \( \phi \) such that \( \vec{F}=\nabla \phi \). (g) Show that the vector field \( \vec{F}=\left(e^{x} \cos y+y z\right) \vec{i}+\left(x z-e^{x} \sin y\right) \vec{j}+(x y+z) \vec{k} \) is conservative by (3 marks) finding a potential for it.