Answer :
Alright, let's go through each part of the question step-by-step.
### Q1: Integrate the following with respect to [tex]\( x \)[/tex]
#### a) [tex]\( \int \frac{1}{x^3} \, dx \)[/tex]
1. Rewrite the integrand:
[tex]\[ \frac{1}{x^3} = x^{-3} \][/tex]
2. Use the power rule for integration, which states:
[tex]\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad \text{for} \quad n \neq -1 \][/tex]
3. Apply the power rule:
[tex]\[ \int x^{-3} \, dx = \frac{x^{-3+1}}{-3+1} + C = \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C \][/tex]
So, the integral is:
[tex]\[ \int \frac{1}{x^3} \, dx = -\frac{1}{2x^2} + C \][/tex]
#### b) [tex]\( \int 6 \sin(3x) \, dx \)[/tex]
1. Recognize that 6 is a constant and can be factored out:
[tex]\[ 6 \int \sin(3x) \, dx \][/tex]
2. Use the substitution method. Let [tex]\( u = 3x \)[/tex], then [tex]\( du = 3 \, dx \)[/tex] or [tex]\( dx = \frac{du}{3} \)[/tex].
3. Substitute [tex]\( u \)[/tex] and [tex]\( dx \)[/tex] into the integral:
[tex]\[ 6 \int \sin(u) \cdot \frac{du}{3} = 2 \int \sin(u) \, du \][/tex]
4. Integrate [tex]\( \sin(u) \)[/tex]:
[tex]\[ 2 \int \sin(u) \, du = -2 \cos(u) + C \][/tex]
5. Substitute back [tex]\( u = 3x \)[/tex]:
[tex]\[ -2 \cos(3x) + C \][/tex]
So, the integral is:
[tex]\[ \int 6 \sin(3x) \, dx = -2 \cos(3x) + C \][/tex]
### Q2: Evaluate the following integrals:
#### a) [tex]\( \int_0^1 \frac{x}{\sqrt{x^2+1}} \, dx \)[/tex]
1. Set up the integral with the given limits:
[tex]\[ \int_0^1 \frac{x}{\sqrt{x^2+1}} \, dx \][/tex]
2. Use the substitution method. Let [tex]\( u = x^2 + 1 \)[/tex], then [tex]\( du = 2x \, dx \)[/tex] or [tex]\( \frac{du}{2} = x \, dx \)[/tex].
3. Change the limits of integration. When [tex]\( x = 0 \)[/tex], [tex]\( u = 1 \)[/tex]. When [tex]\( x = 1 \)[/tex], [tex]\( u = 2 \)[/tex].
4. Substitute [tex]\( u \)[/tex] and [tex]\( dx \)[/tex] into the integral:
[tex]\[ \int_1^2 \frac{1}{\sqrt{u}} \cdot \frac{du}{2} = \frac{1}{2} \int_1^2 u^{-\frac{1}{2}} \, du \][/tex]
5. Integrate [tex]\( u^{-\frac{1}{2}} \)[/tex]:
[tex]\[ \frac{1}{2} \left[ 2 u^{\frac{1}{2}} \right]_1^2 = \left[ u^{\frac{1}{2}} \right]_1^2 \][/tex]
6. Evaluate the expression from 1 to 2:
[tex]\[ 2^{\frac{1}{2}} - 1^{\frac{1}{2}} = \sqrt{2} - 1 \][/tex]
So, the evaluated integral is:
[tex]\[ \int_0^1 \frac{x}{\sqrt{x^2+1}} \, dx = \sqrt{2} - 1 \][/tex]
#### b) [tex]\( \int_0^1 x^2 e^3 \, dx \)[/tex]
1. Recognize that [tex]\( e^3 \)[/tex] is a constant and can be factored out:
[tex]\[ e^3 \int_0^1 x^2 \, dx \][/tex]
2. Integrate [tex]\( x^2 \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \int x^2 \, dx = \frac{x^3}{3} + C \][/tex]
3. Evaluate the integral from 0 to 1:
[tex]\[ \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \][/tex]
4. Multiply by the constant [tex]\( e^3 \)[/tex]:
[tex]\[ e^3 \cdot \frac{1}{3} = \frac{e^3}{3} \][/tex]
So, the evaluated integral is:
[tex]\[ \int_0^1 x^2 e^3 \, dx = \frac{e^3}{3} \][/tex]
### Summary of Results:
1. [tex]\( \int \frac{1}{x^3} \, dx = -\frac{1}{2x^2} + C \)[/tex]
2. [tex]\( \int 6 \sin(3x) \, dx = -2 \cos(3x) + C \)[/tex]
3. [tex]\( \int_0^1 \frac{x}{\sqrt{x^2+1}} \, dx = \sqrt{2} - 1 \)[/tex]
4. [tex]\( \int_0^1 x^2 e^3 \, dx = \frac{e^3}{3} \)[/tex]
### Q1: Integrate the following with respect to [tex]\( x \)[/tex]
#### a) [tex]\( \int \frac{1}{x^3} \, dx \)[/tex]
1. Rewrite the integrand:
[tex]\[ \frac{1}{x^3} = x^{-3} \][/tex]
2. Use the power rule for integration, which states:
[tex]\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad \text{for} \quad n \neq -1 \][/tex]
3. Apply the power rule:
[tex]\[ \int x^{-3} \, dx = \frac{x^{-3+1}}{-3+1} + C = \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C \][/tex]
So, the integral is:
[tex]\[ \int \frac{1}{x^3} \, dx = -\frac{1}{2x^2} + C \][/tex]
#### b) [tex]\( \int 6 \sin(3x) \, dx \)[/tex]
1. Recognize that 6 is a constant and can be factored out:
[tex]\[ 6 \int \sin(3x) \, dx \][/tex]
2. Use the substitution method. Let [tex]\( u = 3x \)[/tex], then [tex]\( du = 3 \, dx \)[/tex] or [tex]\( dx = \frac{du}{3} \)[/tex].
3. Substitute [tex]\( u \)[/tex] and [tex]\( dx \)[/tex] into the integral:
[tex]\[ 6 \int \sin(u) \cdot \frac{du}{3} = 2 \int \sin(u) \, du \][/tex]
4. Integrate [tex]\( \sin(u) \)[/tex]:
[tex]\[ 2 \int \sin(u) \, du = -2 \cos(u) + C \][/tex]
5. Substitute back [tex]\( u = 3x \)[/tex]:
[tex]\[ -2 \cos(3x) + C \][/tex]
So, the integral is:
[tex]\[ \int 6 \sin(3x) \, dx = -2 \cos(3x) + C \][/tex]
### Q2: Evaluate the following integrals:
#### a) [tex]\( \int_0^1 \frac{x}{\sqrt{x^2+1}} \, dx \)[/tex]
1. Set up the integral with the given limits:
[tex]\[ \int_0^1 \frac{x}{\sqrt{x^2+1}} \, dx \][/tex]
2. Use the substitution method. Let [tex]\( u = x^2 + 1 \)[/tex], then [tex]\( du = 2x \, dx \)[/tex] or [tex]\( \frac{du}{2} = x \, dx \)[/tex].
3. Change the limits of integration. When [tex]\( x = 0 \)[/tex], [tex]\( u = 1 \)[/tex]. When [tex]\( x = 1 \)[/tex], [tex]\( u = 2 \)[/tex].
4. Substitute [tex]\( u \)[/tex] and [tex]\( dx \)[/tex] into the integral:
[tex]\[ \int_1^2 \frac{1}{\sqrt{u}} \cdot \frac{du}{2} = \frac{1}{2} \int_1^2 u^{-\frac{1}{2}} \, du \][/tex]
5. Integrate [tex]\( u^{-\frac{1}{2}} \)[/tex]:
[tex]\[ \frac{1}{2} \left[ 2 u^{\frac{1}{2}} \right]_1^2 = \left[ u^{\frac{1}{2}} \right]_1^2 \][/tex]
6. Evaluate the expression from 1 to 2:
[tex]\[ 2^{\frac{1}{2}} - 1^{\frac{1}{2}} = \sqrt{2} - 1 \][/tex]
So, the evaluated integral is:
[tex]\[ \int_0^1 \frac{x}{\sqrt{x^2+1}} \, dx = \sqrt{2} - 1 \][/tex]
#### b) [tex]\( \int_0^1 x^2 e^3 \, dx \)[/tex]
1. Recognize that [tex]\( e^3 \)[/tex] is a constant and can be factored out:
[tex]\[ e^3 \int_0^1 x^2 \, dx \][/tex]
2. Integrate [tex]\( x^2 \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \int x^2 \, dx = \frac{x^3}{3} + C \][/tex]
3. Evaluate the integral from 0 to 1:
[tex]\[ \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \][/tex]
4. Multiply by the constant [tex]\( e^3 \)[/tex]:
[tex]\[ e^3 \cdot \frac{1}{3} = \frac{e^3}{3} \][/tex]
So, the evaluated integral is:
[tex]\[ \int_0^1 x^2 e^3 \, dx = \frac{e^3}{3} \][/tex]
### Summary of Results:
1. [tex]\( \int \frac{1}{x^3} \, dx = -\frac{1}{2x^2} + C \)[/tex]
2. [tex]\( \int 6 \sin(3x) \, dx = -2 \cos(3x) + C \)[/tex]
3. [tex]\( \int_0^1 \frac{x}{\sqrt{x^2+1}} \, dx = \sqrt{2} - 1 \)[/tex]
4. [tex]\( \int_0^1 x^2 e^3 \, dx = \frac{e^3}{3} \)[/tex]
