A rate of change must be positive

Mathematics

1 answer:

Veronika [31]3 years ago

5 0

Answer:

A rate of change can be positive or negative.

Step-by-step explanation:

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$\\\\\ \textbf{a)}\\\\~~~\displaystyle \int (6x- \sin 3x) ~ dx\\\\=6\displaystyle \int x ~ dx - \displaystyle \int \sin 3x ~ dx\\\\=6 \cdot \dfrac{x^2}2 - \dfrac 13 (- \cos 3x) +C~~~~~~~~~~~;\left[\displaystyle \int x^n~ dx = \dfrac{x^{n+1}}{n+1}+C,~~~n \neq -1\right]\\\\ =3x^2 +\dfrac{\cos 3x}3 +C~~~~~~~~~~~~~~~~~~~~;\left[\displaystyle \int \sin (mx) ~dx = -\dfrac 1m ~ (\cos mx)+C \right]\\$

$\textbf{b)}\\\\~~~~\displaystyle \int(3e^{-2x} +\cos (0.5 x)) dx\\\\=3\displaystyle \int e^{-2x} ~dx+ \displaystyle \int \cos(0.5 x) ~dx\\\\\\=-\dfrac 32 e^{-2x} + \dfrac 1{0.5} \sin (0.5 x) +C~~~~~~~~~~~~~~;\left[\displaystyle \int e^{mx}~dx = \dfrac 1m e^{mx} +C \right]\\\\\\=-\dfrac 32 e^{-2x} + 2 \sin(0.5 x) +C~~~~~~~~~~~~~~~~~;\left[\displaystyle \int \cos(mx)~ dx = \dfrac 1m \sin(mx) +C\right]\\\\\\=-1.5e^{-2x} +2\sin(0.5x) +C$

$\textbf{a)}\\\\y = \displaystyle \int \cos(x+5) ~ dx\\\\\text{Let,}\\\\~~~~~~~u = x+5\\\\\implies \dfrac{du}{dx} = 1+0~~~~~~;[\text{Differentiate both sides.}]\\\\\implies \dfrac{du}{dx} = 1\\\\\implies du = dx\\\\\text{Now,}\\\\y= \displaystyle \int \cos u ~ du\\\\~~~= \sin u +C\\\\~~~=\sin(x+5) + C$

$\textbf{b)}\\\\y = \displaystyle \int 2(5x-3)^4 dx\\\\\text{Let,}\\~~~~~~~~u = 5x-3\\\\\implies \dfrac{du}{dx} = 5~~~~~~~~~~;[\text{Differentiate both sides}]\\\\\implies dx = \dfrac{du}5\\\\\text{Now,}\\\\y = 2\cdot \dfrac 1 5 \displaystyle \int u^4 ~ du\\\\\\~~=\dfrac 25 \cdot \dfrac{u^{4+1}}{4+1} +C\\\\\\~~=\dfrac 25 \cdot \dfrac{u^5}5+C\\\\\\~~=\dfrac{2u^5}{25}+C\\\\\\~~=\dfrac{2(5x-3)^5}{25}+C$

$\textbf{a)}\\\\y = \displaystyle \int xe^{3x} dx\\\\\text{We know that,}\\\\ \displaystyle \int (uv) ~dx = u \displaystyle \int v ~ dx - \displaystyle \int \left[ \dfrac{du}{dx} \displaystyle \int ~ v ~ dx \right]~ dx\\\\\text{Let}, u =x~ \text{and}~ v=e^{3x} .\\\\y= \displaystyle \int xe^{3x} ~dx\\\\\\~~= x\displaystyle \int e^{3x} ~ dx - \displaystyle \int \left[\dfrac{d}{dx}(x) \displaystyle \int e^{3x}~ dx \right]~ dx\\\\\\$

$=x\displaystyle \int e^{3x}~ dx - \displaystyle \dfrac 13 \int \left(e^{3x} \right)~ dx\\\\\\=\dfrac{xe^{3x}}3 - \dfrac 13 \cdot \dfrac{ e^{3x}}3+C\\\\\\= \dfrac{xe^{3x}}{3}- \dfrac{e^{3x}}{9}+C\\\\\\=\dfrac{3xe^{3x}}{9}- \dfrac{e^{3x}}9 + C\\\\\\= \dfrac 19e^{3x}(3x-1)+C$