Three Intergral Problems

**Kayla_N** · April 13th 2009, 03:57 PM

1. Use the First Fundamental Theorem of Calculus to find the derivative of

=???

2. Consider the function

.

In this problem you will calculate

by using the definition

The summation inside the brackets is

which is the Riemann sum where the sample points are chosen to be the right-hand endpoints of each subinterval. Calculate

for

on the interval

and write your answer as a function of

without any summation signs.

???

???

( So on this one I did: f(Xk)=f(x)= x^2/2+7) then plug in (2/n *k) for x and got 4/2n^2*k^2+7.
then (4/2n^2*k^2+7) * 2/n and got (8/2n^3*k^2+14/n).
8/2n^3*(2n^3+3n^2+n)= 16n^3+24n^2+8n/12n^3 + 14 answer. did i do something wrong?
For the second part: i took the antiderivative of

and evaluate from 0 to 3. the answer i got is (27/6+7). Help me out thanks.

3.

and

???
The absolute maximum of

occurs when

??? and is the value =????

**Deadstar** · April 13th 2009, 04:52 PM

For the first on note that $tan(t) = \frac{sin(t)}{cos(t)}$ , and since the derivative of cos(t) is -sin(t) we can say the derivative of $\frac{sin(t)}{cos(t)}$ is $-ln(cos(t))$ .

So your integral becomes... $\int_5^{\frac{1}{x}}8tan(t)dt = \int_5^{\frac{1}{x}}8\frac{sin(t)}{cos(t)}(t)dt = -8ln(cos(\frac{1}{x})) - (-8ln(cos(5)))$ .

**mr fantastic** · April 13th 2009, 07:43 PM

Originally Posted by Deadstar

For the first on note that $tan(t) = \frac{sin(t)}{cos(t)}$ , and since the derivative of cos(t) is -sin(t) we can say the derivative of $\frac{sin(t)}{cos(t)}$ is $-ln(cos(t))$ .

So your integral becomes... $\int_5^{\frac{1}{x}}8tan(t)dt = \int_5^{\frac{1}{x}}8\frac{sin(t)}{cos(t)}(t)dt = -8ln(cos(\frac{1}{x})) - (-8ln(cos(5)))$ .

I think what's needed here (if it's to be done in the way the question instructs) is to first make the substitution $u = \frac{1}{x}$ :

$y = \int_{5}^{1/x} 8 \tan (t) \, dt = \int_{5}^{u} 8 \tan (t) \, dt$ .

Now use the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ .

From the Fundamental Theorem of Calculus $\frac{dy}{du} = 8 \tan (u)$ and obviously $\frac{du}{dx} = - \frac{1}{x^2}$ .

Therefore $h'(x) = \frac{dy}{dx} = - 8 \tan (u) \cdot \frac{1}{x^2} = - 8 \tan \left( \frac{1}{x} \right) \cdot \frac{1}{x^2} = - \frac{8}{x^2} \tan \left( \frac{1}{x} \right)$ .

**Kayla_N** · April 14th 2009, 11:33 AM

Can someone please look at number 2. Did i do something wrong or missing something? I cant figured out. Also i need help on 3 also. Thanks

Thread: Three Intergral Problems

LinkBack

Thread Tools

Display

Three Intergral Problems

Similar Math Help Forum Discussions

Intergral

intergral of sin^4(x)cos^2(x)

Intergral Problems!!!

fundamental theorem of calculus definite intergral problems

Intergral

Search Tags