# Thread: hi plz help in this integral homework?????????

1. ## hi plz help in this integral homework?????????

hello all
if u can plz help me with this homework i attached it here

thx 4 all

2. Originally Posted by eman.kh
hello all
if u can plz help me with this homework i attached it here

thx 4 all
$f(x) = 6x - e^{1 - 2x}$

a)
$A = \int_1^5 (6x - e^{1 - 2x})dx$
and
$W = \frac{1}{4 - 1} \int_1^4 (6x - e^{1 - 2x})dx$

To do either of these we need
$\int (6x - e^{1 - 2x})dx$

Let's rewrite this a little:
$= \int (6x - e^1 \cdot e^{-2x})dx$

$= \int 6x~dx - e \int e^{-2x}~dx$

The first integral is elementary, for the second one use the substitution u = -2x and the fact that $\int e^u ~ du = e^u + C$.

b)
$f^{\prime}(x) = 6 - e^{1 - 2x} \cdot -2 = 6 + 2e^{1 - 2x}$

$f^{\prime \prime}(x) = 2e^{1 - 2x} \cdot -2 = -4e^{1 - 2x}$

So just plug these into the expression $f^{\prime \prime} + 2f^{\prime} - 12$ and show that it has a value of 0.

c)
Concavity is determined by the second derivative. So to get the inflection points solve $f^{\prime \prime}(x) = 0$ for x. (Hint: $e^{x} \geq 0$ for all x, so are there any values of x such that the second derivative is 0?)

To determine the concavity of the function, if $f^{\prime \prime}(x) > 0$ the function is concave up, if $f^{\prime \prime}(x) < 0$ the function is concave down.

-Dan

3. really thanks alot 4 helping me,idon't know how to thank u .

thanks.

4. Originally Posted by eman.kh
really thanks alot 4 helping me,idon't know how to thank u .

thanks.
That is thanks enough for me!

-Dan