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1

he Laplace Transform is an integral transform that converts functions from the time domain to the complex frequency domain . The transformation of a function into its Laplace Transform is given by:

$F(s)= \int_{0}^{\infty}e^{-st}*f(t) dt$

Where , where and are constants and is the imaginary number.

Give the Laplace Transform of $\frac{\mathrm{d} }{\mathrm{d} t}[f(t)]$ .

Explanation

The Laplace Transform of the derivative is given by:

$\int_{0}^{\infty}e^{-st}*f'(t) dt$

Using integration by parts,

$\int_{0}^{\infty} UdV= [UV]_{0}^{\infty}-\int_{0}^{\infty} VdU$

Let and $U=e^{-st}$

$[e^{-st}*f(t) ]_{0}^{\infty}-\int_{0}^{\infty}(-s)e^{-st}*f(t) dt$

The first term becomes

and the second term becomes

The Laplace Transform therefore becomes:

2

$\int_{0}^{2}x^2-\frac{1}{2}xdx$

$\frac{5}{3}$

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{4}{3}$

$\frac{10}{3}$

Explanation

First, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{x^3}{3}-\frac{1}{2}(\frac{x^2}{2})=\frac{x^3}{3}-\frac{x^2}{4}$

Now, evaluate at 2 and then 0. Subtract the results:

$(\frac{2^3}{3})-\frac{2^2}{4}-0=$

Simplify to get your answer:

$\frac{8}{3}-\frac{3}{3}=\frac{5}{3}$

3

$\int_{0}^{2}x^2-\frac{1}{2}xdx$

$\frac{5}{3}$

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{4}{3}$

$\frac{10}{3}$

Explanation

First, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{x^3}{3}-\frac{1}{2}(\frac{x^2}{2})=\frac{x^3}{3}-\frac{x^2}{4}$

Now, evaluate at 2 and then 0. Subtract the results:

$(\frac{2^3}{3})-\frac{2^2}{4}-0=$

Simplify to get your answer:

$\frac{8}{3}-\frac{3}{3}=\frac{5}{3}$

4

$\int_{0}^{2}x^2-\frac{1}{2}xdx$

$\frac{5}{3}$

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{4}{3}$

$\frac{10}{3}$

Explanation

First, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{x^3}{3}-\frac{1}{2}(\frac{x^2}{2})=\frac{x^3}{3}-\frac{x^2}{4}$

Now, evaluate at 2 and then 0. Subtract the results:

$(\frac{2^3}{3})-\frac{2^2}{4}-0=$

Simplify to get your answer:

$\frac{8}{3}-\frac{3}{3}=\frac{5}{3}$

5

Evaluate.

$\int 12 + x \ dx$

$F(x) = 12x + \frac{x^2}{2} + C$

$F(x) = \frac{6x}{2} + \frac{x^2}{2} +C$

Answer not listed.

Explanation

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of

If then

If then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then

If then

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, .

The antiderivative is $F(x) = 12x + \frac{x^2}{2} +C$ .

6

Evaluate.

$\int 12 + x \ dx$

$F(x) = 12x + \frac{x^2}{2} + C$

$F(x) = \frac{6x}{2} + \frac{x^2}{2} +C$

Answer not listed.

Explanation

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of

If then

If then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then

If then

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, .

The antiderivative is $F(x) = 12x + \frac{x^2}{2} +C$ .

7

Evaluate.

$\int 12 + x \ dx$

$F(x) = 12x + \frac{x^2}{2} + C$

$F(x) = \frac{6x}{2} + \frac{x^2}{2} +C$

Answer not listed.

Explanation

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of

If then

If then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then

If then

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, .

The antiderivative is $F(x) = 12x + \frac{x^2}{2} +C$ .

8

$\int_{0}^{4}2x^2-x+3dx$

$\frac{140}{3}$

$\frac{137}{2}$

$\frac{140}{4}$

$\frac{131}{3}$

Explanation

First, integrate the expression. Remember to raise the exponent by 1 and then also put that result on the denominator.

Therefore, after integrating, it should look like:

$\frac{2x^3}{3}-\frac{x^2}{2}+3x$ .

Then, first evaluate at 4 and then 0.

Subtract the results:

$(\frac{2(64)}{3}-8+12)-0=\frac{128}{3}+4=\frac{140}{3}$ .

9

$\int_{0}^{4}2x^2-x+3dx$

$\frac{140}{3}$

$\frac{137}{2}$

$\frac{140}{4}$

$\frac{131}{3}$

Explanation

First, integrate the expression. Remember to raise the exponent by 1 and then also put that result on the denominator.

Therefore, after integrating, it should look like:

$\frac{2x^3}{3}-\frac{x^2}{2}+3x$ .

Then, first evaluate at 4 and then 0.

Subtract the results:

$(\frac{2(64)}{3}-8+12)-0=\frac{128}{3}+4=\frac{140}{3}$ .

10

$\int_{0}^{4}2x^2-x+3dx$

$\frac{140}{3}$

$\frac{137}{2}$

$\frac{140}{4}$

$\frac{131}{3}$

Explanation

First, integrate the expression. Remember to raise the exponent by 1 and then also put that result on the denominator.

Therefore, after integrating, it should look like:

$\frac{2x^3}{3}-\frac{x^2}{2}+3x$ .

Then, first evaluate at 4 and then 0.

Subtract the results:

$(\frac{2(64)}{3}-8+12)-0=\frac{128}{3}+4=\frac{140}{3}$ .

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