The constant of integration, \(c\) is added as a part of indefinite integral for a function. For example,

\(\int 5 \, dx=5x+c\).

If \(f(x)\) and \(g(x)\) are functions, then

\(\int [f(x)\pm g(x)] \, dx =\int f(x) \, dx \pm \int g(x) \, dx\).

Substitution method can be use to solve any functions in the form of \((ax+b)^n\). Then,

​\(\int (ax+b)^n \, dx=\dfrac{(ax+b)^{n+1}}{a(n+1)}+c\),

​where \(a\) and \(b\) are constants, \(n\) is an integer and \(n \neq -1\).

Integrate each of the following with the respect to \(x\):

(a) \(-0.5\),
(b) \(\int \dfrac{2}{x^2} \, dx\).

(a)

\(\int -0.5 \ dx = -0.5x +c\).

(b)

\(\begin{aligned} \int \dfrac{2}{x^2} \ dx &= 2 \int x^{-2} \ dx\\\\ &= 2 \begin{pmatrix} \dfrac{x^{-2+1}}{-2+1} \end{pmatrix} +c\\\\ &= -2x^{-1} +c\\\\ &= -\dfrac{2}{x} +c. \end{aligned}\)

Find the integral for the following

\(\int (x-2)(x+6) \, dx\)

\(\begin{aligned} \int (x-2)(x+6) \ dx &= \int (x^2 +4x - 12) \ dx\\\\ &= \int x^2 \ dx + \int 4x \ dx-\int12 \ dx\\\\ &= \dfrac{x^3}{3} + \dfrac{4x^2}{2} - 12x +c\\\\ &= \dfrac{x^3}{3} + 2x^2- 12x +c. \end{aligned}\)

(a) By using the substitution method, find the indefinite integral for the following

\(\int \sqrt{5x+2} \, dx\)

(b) The gradient function of a curve at point \((x,y)\) is given by

\(\dfrac{dy}{dx} = 15x^2 + 4x- 3\)

If the curve passes through the point \((-1, 2)\),  find the equation of the curve.

(a)

Let \(u=5x+2\), so,

\(\begin{aligned} \dfrac{du}{dx} &=5\\\\ dx&= \dfrac{du}{5} \end{aligned}\)

\(\begin{aligned} \int \sqrt{5x+2} \ dx &= \int \dfrac{\sqrt u}{5} \ du\\\\ &= \int \dfrac{u^{\frac{1}{2}}}{5} \ du\\\\ &=\dfrac{2}{15}u^{\frac{3}{2}} +c\\\\ &= \dfrac{2}{15}(5x+2)^{\frac{3}{2}}+c. \end{aligned}\)

(b)

Given \(\dfrac{dy}{dx} = 15x^2 + 4x- 3\).

Then,

\(\begin{aligned} y&=\int (15x^2 +4x-3) \ dx\\\\ y&=5x^3 +2x^2-3x+c. \end{aligned}\)

When \(x=-1\) and \(y=2\),

\(\begin{aligned} 2&=5(-1)^3+2(-1)^2-3(-1)+c\\ c&=2 .\end{aligned}\)

Thus, the equation of the curve is:

\(y=5x^3 +2x^2-3x+2\).

The constant of integration, \(c\) is added as a part of indefinite integral for a function. For example,

\(\int 5 \, dx=5x+c\).

If \(f(x)\) and \(g(x)\) are functions, then

\(\int [f(x)\pm g(x)] \, dx =\int f(x) \, dx \pm \int g(x) \, dx\).

Substitution method can be use to solve any functions in the form of \((ax+b)^n\). Then,

​\(\int (ax+b)^n \, dx=\dfrac{(ax+b)^{n+1}}{a(n+1)}+c\),

​where \(a\) and \(b\) are constants, \(n\) is an integer and \(n \neq -1\).

Integrate each of the following with the respect to \(x\):

(a) \(-0.5\),
(b) \(\int \dfrac{2}{x^2} \, dx\).

(a)

\(\int -0.5 \ dx = -0.5x +c\).

(b)

\(\begin{aligned} \int \dfrac{2}{x^2} \ dx &= 2 \int x^{-2} \ dx\\\\ &= 2 \begin{pmatrix} \dfrac{x^{-2+1}}{-2+1} \end{pmatrix} +c\\\\ &= -2x^{-1} +c\\\\ &= -\dfrac{2}{x} +c. \end{aligned}\)

Find the integral for the following

\(\int (x-2)(x+6) \, dx\)

\(\begin{aligned} \int (x-2)(x+6) \ dx &= \int (x^2 +4x - 12) \ dx\\\\ &= \int x^2 \ dx + \int 4x \ dx-\int12 \ dx\\\\ &= \dfrac{x^3}{3} + \dfrac{4x^2}{2} - 12x +c\\\\ &= \dfrac{x^3}{3} + 2x^2- 12x +c. \end{aligned}\)

(a) By using the substitution method, find the indefinite integral for the following

\(\int \sqrt{5x+2} \, dx\)

(b) The gradient function of a curve at point \((x,y)\) is given by

\(\dfrac{dy}{dx} = 15x^2 + 4x- 3\)

If the curve passes through the point \((-1, 2)\),  find the equation of the curve.

(a)

Let \(u=5x+2\), so,

\(\begin{aligned} \dfrac{du}{dx} &=5\\\\ dx&= \dfrac{du}{5} \end{aligned}\)

\(\begin{aligned} \int \sqrt{5x+2} \ dx &= \int \dfrac{\sqrt u}{5} \ du\\\\ &= \int \dfrac{u^{\frac{1}{2}}}{5} \ du\\\\ &=\dfrac{2}{15}u^{\frac{3}{2}} +c\\\\ &= \dfrac{2}{15}(5x+2)^{\frac{3}{2}}+c. \end{aligned}\)

(b)

Given \(\dfrac{dy}{dx} = 15x^2 + 4x- 3\).

Then,

\(\begin{aligned} y&=\int (15x^2 +4x-3) \ dx\\\\ y&=5x^3 +2x^2-3x+c. \end{aligned}\)

When \(x=-1\) and \(y=2\),

\(\begin{aligned} 2&=5(-1)^3+2(-1)^2-3(-1)+c\\ c&=2 .\end{aligned}\)

Thus, the equation of the curve is:

\(y=5x^3 +2x^2-3x+2\).