Post #25

So I'm going to explain some of the questions I missed on the non-calculator portion because some of yall may have missed them too.

An equation of the line normal to the graph of y=the square root of (3x^2+2x) at (2,4) is
A. -4x+y=20 B. 4x+7y=20 C. -7x+4y=2 D. 7x+4y=30 E. 4x+7y=30

To find an equation of a normal line, you need a point and a perpendicular slope. To find the slope, take the derivative of the equation and plug in your point. The perpendicular slope will be the negative reciprocal of the slope.
Derivative:
(3x^2+2x)^1/2
1/2(3x^2+2x)^-1/2 (6x+2)
6x+2/ 2 (3x^2 +2x)^1/2
Plug in:
6(2) + 2 / 2(3(2)^2 + 2(2)) ^1/2
14/ 8
m= 7/4
the perpendicular slope = -4/7

Now that you have your perpendicular slope, you plug into point-slope formula.
y-4 = -4/7 (x-2)
Since that answer does not appear in the choice, you need to solve the equation for an answer they give. For the equation, multiply everything by seven to get rid of the fraction first then solve
7y-28 = -4 ( x-2)
7y-28 =-4x+8
4x+7y -36 = 0
4x+7y = 36
Therefore, the answer is D.

The integral of (e^3lnx + e^3x) dx=
The key word in this problem is the integral sign. Looking at the first term, you know the ln and e cancel leaving you with x^3.
The problem is now the integral of x^3 + e^3x now integrate.
The integral of x^3 is 1/4 x^4 and the integral of e^3x is 1/3 e^3x
So the answer will be x^4/4 + e^3x/3 + c.

The average value of the function (x-1)^2 on the interval from x=1 to x=5 is
A. -16/3 B. 16/3 C. 64/3 D. 66/3 E. 256/3
To find the average value of a function, you do 1/b-a times the integral of the function on the interval given.
1/ 5-1 integral (x-1)^2 on [1,5]
To integrate this function, you first have to factor it out.
x^2 - 2x +1
Integrate: 1/3 x^3 - x^2 + x on [1,5]
Next step is to plug in and multiply by 1/b-a
1/4 [1/3 (5)^3 -(5)^2 +5] - [1/3(1)^3 - (1)^2 +1 ]
1/4 [ 1/3 (125) - 25 +5 ] - [1/3 -1 +1]
1/4 ( 125/3 - 25 + 5 - 1/3 )
1/4 (64/3 )
64/12 = 16/3 So the answer is B.

I have questions on 30, 35, 39, 42, and 44 on the calculator portion of the last AP.