Need help with MATH problem....please thanks in advance!

[mjvskb24vs23]

nx7=70

 

[po2345]

Originally Posted by jordan723

^^^^I noticed the pattern but so what is the formula????????????????????

you have to use n such as (n*2+4) something like that

N(3+N).

PO

 

[po2345]

Originally Posted by jordan723

^^^^I noticed the pattern but so what is the formula????????????????????

you have to use n such as (n*2+4) something like that

N(3+N).

PO

 

[dr fonclbrics]

Originally Posted by Name I Koop

# of tooothpicks = N x (3+N) = 3N + N^2

/thread

 

[jfmartimcdandruff]

Suppose f(x) and g(x) are two continuously differentiable functions. The product rule states

(f(x)g(x))' = f(x) g'(x) + f'(x) g(x)\!

Integrating both sides gives

f(x) g(x) = \int f(x) g'(x)\, dx + \int f'(x) g(x)\, dx\!

Rearranging terms

\int f(x) g'(x)\, dx = f(x) g(x) - \int f'(x) g(x)\, dx\!

From the above one can derive the integration by parts rule, which states that, given an interval with endpoints a and b,

\int_a^b f(x) g'(x)\, dx = \left[ f(x) g(x) ight]_a^b - \int_a^b f'(x) g(x)\, dx\!

with the common notation

\left[ f(x) g(x) ight]_a^b = f(b) g(b) - f(a) g(a).\!

The rule is shown to be true by using the product rule for derivatives and the fundamental theorem of calculus. Thus

\begin{align} f(b)g(b) - f(a)g(a) & = \int_a^b \frac{d}{dx} ( f(x) g(x) )\, dx \\ & = \int_a^b f'(x) g(x) \, dx + \int_a^b f(x) g'(x)\, dx. \end{align}

In the traditional calculus curriculum, the rule is often stated using indefinite integrals in the form

\int f(x) g'(x)\, dx = f(x) g(x) - \int f'(x) g(x)\, dx\!

or, if u = f(x), v = g(x) and the differentials du = f ′(x) dx and dv = g′(x) dx, then it is in the form most often seen:

\int u\, dv=uv-\int v\, du.\!

This formula can be interpreted to mean that the area under the graph of a function u(v) is the same as the area of the rectangle u v minus the area above the graph.

The original integral contains the derivative of g; in order to be able to apply the rule, the antiderivative g must be found, and then the resulting integral ∫g f ′ dx must be evaluated.

One can also formulate a discrete analogue for sequences, called summation by parts.

An alternative notation has the advantage that the factors of the original expression are identified as f(x) and g(x):

\int_a^b f(x) g(x)\, dx = \left[ f(x) \int g(x)\, dx ight]_a^b - \int_a^b \left ( \int g(x)\, dx ight )\, f'(x) \, dx.\!

This formula is valid whenever f(x) is continuously differentiable and g(x) is continuous.

More general formulations of integration by parts exist for the Riemann–Stieltjes integral and Lebesgue–Stieltjes integral.

More complicated forms of the rule are also valid:

\int u v\, dw = u v w - \int u w\, dv - \int v w\, du.\!

 

[dr fonclbrics]

Originally Posted by Name I Koop

# of tooothpicks = N x (3+N) = 3N + N^2

/thread

 

[jfmartimcdandruff]

Suppose f(x) and g(x) are two continuously differentiable functions. The product rule states

(f(x)g(x))' = f(x) g'(x) + f'(x) g(x)\!

Integrating both sides gives

f(x) g(x) = \int f(x) g'(x)\, dx + \int f'(x) g(x)\, dx\!

Rearranging terms

\int f(x) g'(x)\, dx = f(x) g(x) - \int f'(x) g(x)\, dx\!

From the above one can derive the integration by parts rule, which states that, given an interval with endpoints a and b,

\int_a^b f(x) g'(x)\, dx = \left[ f(x) g(x) ight]_a^b - \int_a^b f'(x) g(x)\, dx\!

with the common notation

\left[ f(x) g(x) ight]_a^b = f(b) g(b) - f(a) g(a).\!

The rule is shown to be true by using the product rule for derivatives and the fundamental theorem of calculus. Thus

\begin{align} f(b)g(b) - f(a)g(a) & = \int_a^b \frac{d}{dx} ( f(x) g(x) )\, dx \\ & = \int_a^b f'(x) g(x) \, dx + \int_a^b f(x) g'(x)\, dx. \end{align}

In the traditional calculus curriculum, the rule is often stated using indefinite integrals in the form

\int f(x) g'(x)\, dx = f(x) g(x) - \int f'(x) g(x)\, dx\!

or, if u = f(x), v = g(x) and the differentials du = f ′(x) dx and dv = g′(x) dx, then it is in the form most often seen:

\int u\, dv=uv-\int v\, du.\!

This formula can be interpreted to mean that the area under the graph of a function u(v) is the same as the area of the rectangle u v minus the area above the graph.

The original integral contains the derivative of g; in order to be able to apply the rule, the antiderivative g must be found, and then the resulting integral ∫g f ′ dx must be evaluated.

One can also formulate a discrete analogue for sequences, called summation by parts.

An alternative notation has the advantage that the factors of the original expression are identified as f(x) and g(x):

\int_a^b f(x) g(x)\, dx = \left[ f(x) \int g(x)\, dx ight]_a^b - \int_a^b \left ( \int g(x)\, dx ight )\, f'(x) \, dx.\!

This formula is valid whenever f(x) is continuously differentiable and g(x) is continuous.

More general formulations of integration by parts exist for the Riemann–Stieltjes integral and Lebesgue–Stieltjes integral.

More complicated forms of the rule are also valid:

\int u v\, dw = u v w - \int u w\, dv - \int v w\, du.\!

 

[jordan723]

thank you ^^^^^^^ finally.....appreciate it

 

[jordan723]

thank you ^^^^^^^ finally.....appreciate it

 

[onewearz]

 

[onewearz]

 

[kilojules64]

^ It took 6 minutes, what do you mean finally

 

[kilojules64]

^ It took 6 minutes, what do you mean finally

 

[superflyinchopstickninja]

Originally Posted by jordan723

thank you ^^^^^^^ finally.....appreciate it

the formula was given to you multiple times before that.  honestly, if people give you the answer and youre STILL unable to figure out the work, then go get a tutor.  or go to an easier math class.

 

[superflyinchopstickninja]

Originally Posted by jordan723

thank you ^^^^^^^ finally.....appreciate it

the formula was given to you multiple times before that.  honestly, if people give you the answer and youre STILL unable to figure out the work, then go get a tutor.  or go to an easier math class.

 

[shatterkneesinc]

are u smarter than a 5th grader?

 

[shatterkneesinc]

are u smarter than a 5th grader?

 

[mondaynightraw]

Originally Posted by JFMartiMcDandruff

Suppose f(x) and g(x) are two continuously differentiable functions. The product rule states

(f(x)g(x))' = f(x) g'(x) + f'(x) g(x)\!

Integrating both sides gives

f(x) g(x) = \int f(x) g'(x)\, dx + \int f'(x) g(x)\, dx\!

Rearranging terms

\int f(x) g'(x)\, dx = f(x) g(x) - \int f'(x) g(x)\, dx\!

From the above one can derive the integration by parts rule, which states that, given an interval with endpoints a and b,

\int_a^b f(x) g'(x)\, dx = \left[ f(x) g(x) ight]_a^b - \int_a^b f'(x) g(x)\, dx\!

with the common notation

\left[ f(x) g(x) ight]_a^b = f(b) g(b) - f(a) g(a).\!

The rule is shown to be true by using the product rule for derivatives and the fundamental theorem of calculus. Thus

\begin{align} f(b)g(b) - f(a)g(a) & = \int_a^b \frac{d}{dx} ( f(x) g(x) )\, dx \\ & = \int_a^b f'(x) g(x) \, dx + \int_a^b f(x) g'(x)\, dx. \end{align}

In the traditional calculus curriculum, the rule is often stated using indefinite integrals in the form

\int f(x) g'(x)\, dx = f(x) g(x) - \int f'(x) g(x)\, dx\!

or, if u = f(x), v = g(x) and the differentials du = f ′(x) dx and dv = g′(x) dx, then it is in the form most often seen:

\int u\, dv=uv-\int v\, du.\!

This formula can be interpreted to mean that the area under the graph of a function u(v) is the same as the area of the rectangle u v minus the area above the graph.

The original integral contains the derivative of g; in order to be able to apply the rule, the antiderivative g must be found, and then the resulting integral ∫g f ′ dx must be evaluated.

One can also formulate a discrete analogue for sequences, called summation by parts.

An alternative notation has the advantage that the factors of the original expression are identified as f(x) and g(x):

\int_a^b f(x) g(x)\, dx = \left[ f(x) \int g(x)\, dx ight]_a^b - \int_a^b \left ( \int g(x)\, dx ight )\, f'(x) \, dx.\!

This formula is valid whenever f(x) is continuously differentiable and g(x) is continuous.

More general formulations of integration by parts exist for the Riemann–Stieltjes integral and Lebesgue–Stieltjes integral.

More complicated forms of the rule are also valid:

\int u v\, dw = u v w - \int u w\, dv - \int v w\, du.\!

 

[mondaynightraw]

Originally Posted by JFMartiMcDandruff

Suppose f(x) and g(x) are two continuously differentiable functions. The product rule states

(f(x)g(x))' = f(x) g'(x) + f'(x) g(x)\!

Integrating both sides gives

f(x) g(x) = \int f(x) g'(x)\, dx + \int f'(x) g(x)\, dx\!

Rearranging terms

\int f(x) g'(x)\, dx = f(x) g(x) - \int f'(x) g(x)\, dx\!

From the above one can derive the integration by parts rule, which states that, given an interval with endpoints a and b,

\int_a^b f(x) g'(x)\, dx = \left[ f(x) g(x) ight]_a^b - \int_a^b f'(x) g(x)\, dx\!

with the common notation

\left[ f(x) g(x) ight]_a^b = f(b) g(b) - f(a) g(a).\!

The rule is shown to be true by using the product rule for derivatives and the fundamental theorem of calculus. Thus

\begin{align} f(b)g(b) - f(a)g(a) & = \int_a^b \frac{d}{dx} ( f(x) g(x) )\, dx \\ & = \int_a^b f'(x) g(x) \, dx + \int_a^b f(x) g'(x)\, dx. \end{align}

In the traditional calculus curriculum, the rule is often stated using indefinite integrals in the form

\int f(x) g'(x)\, dx = f(x) g(x) - \int f'(x) g(x)\, dx\!

or, if u = f(x), v = g(x) and the differentials du = f ′(x) dx and dv = g′(x) dx, then it is in the form most often seen:

\int u\, dv=uv-\int v\, du.\!

This formula can be interpreted to mean that the area under the graph of a function u(v) is the same as the area of the rectangle u v minus the area above the graph.

The original integral contains the derivative of g; in order to be able to apply the rule, the antiderivative g must be found, and then the resulting integral ∫g f ′ dx must be evaluated.

One can also formulate a discrete analogue for sequences, called summation by parts.

An alternative notation has the advantage that the factors of the original expression are identified as f(x) and g(x):

\int_a^b f(x) g(x)\, dx = \left[ f(x) \int g(x)\, dx ight]_a^b - \int_a^b \left ( \int g(x)\, dx ight )\, f'(x) \, dx.\!

This formula is valid whenever f(x) is continuously differentiable and g(x) is continuous.

More general formulations of integration by parts exist for the Riemann–Stieltjes integral and Lebesgue–Stieltjes integral.

More complicated forms of the rule are also valid:

\int u v\, dw = u v w - \int u w\, dv - \int v w\, du.\!

 

[spidermachine916]

oooo now i get it....
thanks JFMartiMcDandruff

 

[spidermachine916]

oooo now i get it....
thanks JFMartiMcDandruff

 

[jordanfean23]

Theres actually a website that gives you any answer to any problem...Google it.

 

[jordanfean23]

Theres actually a website that gives you any answer to any problem...Google it.

 

[elcatfisho]

Originally Posted by mondaynightraw

Originally Posted by JFMartiMcDandruff

Suppose f(x) and g(x) are two continuously differentiable functions. The product rule states

(f(x)g(x))' = f(x) g'(x) + f'(x) g(x)\!

Integrating both sides gives

f(x) g(x) = \int f(x) g'(x)\, dx + \int f'(x) g(x)\, dx\!

Rearranging terms

\int f(x) g'(x)\, dx = f(x) g(x) - \int f'(x) g(x)\, dx\!

From the above one can derive the integration by parts rule, which states that, given an interval with endpoints a and b,

\int_a^b f(x) g'(x)\, dx = \left[ f(x) g(x) ight]_a^b - \int_a^b f'(x) g(x)\, dx\!

with the common notation

\left[ f(x) g(x) ight]_a^b = f(b) g(b) - f(a) g(a).\!

The rule is shown to be true by using the product rule for derivatives and the fundamental theorem of calculus. Thus

\begin{align} f(b)g(b) - f(a)g(a) & = \int_a^b \frac{d}{dx} ( f(x) g(x) )\, dx \\ & = \int_a^b f'(x) g(x) \, dx + \int_a^b f(x) g'(x)\, dx. \end{align}

In the traditional calculus curriculum, the rule is often stated using indefinite integrals in the form

\int f(x) g'(x)\, dx = f(x) g(x) - \int f'(x) g(x)\, dx\!

or, if u = f(x), v = g(x) and the differentials du = f ′(x) dx and dv = g′(x) dx, then it is in the form most often seen:

\int u\, dv=uv-\int v\, du.\!

This formula can be interpreted to mean that the area under the graph of a function u(v) is the same as the area of the rectangle u v minus the area above the graph.

The original integral contains the derivative of g; in order to be able to apply the rule, the antiderivative g must be found, and then the resulting integral ∫g f ′ dx must be evaluated.

One can also formulate a discrete analogue for sequences, called summation by parts.

An alternative notation has the advantage that the factors of the original expression are identified as f(x) and g(x):

\int_a^b f(x) g(x)\, dx = \left[ f(x) \int g(x)\, dx ight]_a^b - \int_a^b \left ( \int g(x)\, dx ight )\, f'(x) \, dx.\!

This formula is valid whenever f(x) is continuously differentiable and g(x) is continuous.

More general formulations of integration by parts exist for the Riemann–Stieltjes integral and Lebesgue–Stieltjes integral.

More complicated forms of the rule are also valid:

\int u v\, dw = u v w - \int u w\, dv - \int v w\, du.\!

lol integration by parts. getting tested on that today

 

[elcatfisho]

Originally Posted by mondaynightraw

Originally Posted by JFMartiMcDandruff

Suppose f(x) and g(x) are two continuously differentiable functions. The product rule states

(f(x)g(x))' = f(x) g'(x) + f'(x) g(x)\!

Integrating both sides gives

f(x) g(x) = \int f(x) g'(x)\, dx + \int f'(x) g(x)\, dx\!

Rearranging terms

\int f(x) g'(x)\, dx = f(x) g(x) - \int f'(x) g(x)\, dx\!

From the above one can derive the integration by parts rule, which states that, given an interval with endpoints a and b,

\int_a^b f(x) g'(x)\, dx = \left[ f(x) g(x) ight]_a^b - \int_a^b f'(x) g(x)\, dx\!

with the common notation

\left[ f(x) g(x) ight]_a^b = f(b) g(b) - f(a) g(a).\!

The rule is shown to be true by using the product rule for derivatives and the fundamental theorem of calculus. Thus

\begin{align} f(b)g(b) - f(a)g(a) & = \int_a^b \frac{d}{dx} ( f(x) g(x) )\, dx \\ & = \int_a^b f'(x) g(x) \, dx + \int_a^b f(x) g'(x)\, dx. \end{align}

In the traditional calculus curriculum, the rule is often stated using indefinite integrals in the form

\int f(x) g'(x)\, dx = f(x) g(x) - \int f'(x) g(x)\, dx\!

or, if u = f(x), v = g(x) and the differentials du = f ′(x) dx and dv = g′(x) dx, then it is in the form most often seen:

\int u\, dv=uv-\int v\, du.\!

This formula can be interpreted to mean that the area under the graph of a function u(v) is the same as the area of the rectangle u v minus the area above the graph.

The original integral contains the derivative of g; in order to be able to apply the rule, the antiderivative g must be found, and then the resulting integral ∫g f ′ dx must be evaluated.

One can also formulate a discrete analogue for sequences, called summation by parts.

An alternative notation has the advantage that the factors of the original expression are identified as f(x) and g(x):

\int_a^b f(x) g(x)\, dx = \left[ f(x) \int g(x)\, dx ight]_a^b - \int_a^b \left ( \int g(x)\, dx ight )\, f'(x) \, dx.\!

This formula is valid whenever f(x) is continuously differentiable and g(x) is continuous.

More general formulations of integration by parts exist for the Riemann–Stieltjes integral and Lebesgue–Stieltjes integral.

More complicated forms of the rule are also valid:

\int u v\, dw = u v w - \int u w\, dv - \int v w\, du.\!

lol integration by parts. getting tested on that today