Calculus 1 Help Needed

Calculus Help Needed?

Just a few questions im stuck, new to calculus.

1.) Evaluate the sum
I tried the formula 1/6n(n+1)(2n+1) however it was off by 5 from real value.

20
∑ i^2
i=3

2.) I ended up with (alpha) * ( (n^(n+1)-1) / (n-1) )

n
∑ (alpha) * n^i
i=0

3.) I ended up with -1 + (n+1)^3 think its a telescoping sum

n
∑ ( (i+1)^3 - i^3 )
i=1

4.)

2n
∑ (beta) * i^2
i=n+1

Any hints or working is appreciated. Thanks in advance

Calculus Help Needed?

I just can't figure these two problems out. Help is much appreciated


1. When a camera flash goes off, the batteries immediately begin to recharge the flash's capacitor, which stores electric charge given by the following.

Q(t) = Q0(1 − e^(−t/a))

(The maximum charge capacity is Q0 and t is measured in seconds.)

(a) Find the inverse of this function.

(b) How long does it take to recharge the capacitor to 74% of capacity if a = 5? (Round your answer to one decimal place.)





2. e^(ax) = Ce^(bx), where a ≠ b

x = ?

Calculus help needed?

YOU should certainly see that g ' > 0 and g ' ' = e^(- x²) > 0

Calculus Help needed!!!?

Consider the following function.
f(x) = x ln 3x, a = 1, n = 3, 0.7 ≤ x ≤ 1.3
1. Approximate f by a Taylor polynomial with degree n at the number a. T3(x)=?
2. Use Taylor's Inequality to estimate the accuracy of the approximation
f(x) ≈ Tn(x)
when x lies in the given interval. (Round your answer to four decimal places.)
|R3(x)| ≤ ?

Need Calculus 1 Help?

Subtract from the power, not add.
d/dx [x⁻²] = -2x⁻³

AP Calculus help needed?

Make up a test function and check the pattern

F(x) = x^2
F'(x) = f(x) = 2x

Let's integrate f(x) from 1 to 3. That'd be F(3) - F(1)
Now, let's integrate f(2x) from 1 to 3

F(2x) = (2x)^2
F'(2x) = 2 * (2x)^(2 - 1) * 2 = 4 * 2x = 8x

If we integrate f(2x) from 1 to 3, we'd get

F(2 * 3) - F(2 * 1) = F(6) - F(2)

Is that one of the options?

If we're looking at something that is supposed to be a general case, then all we have to do is pick a specific case and work with it. The general rule that the integral of f(k * x) * dx from x = a to x = b is equal to the integral of f(x) * dx from x = ka to x = kb is confirmed with the specific cases of F(x) and F(2x) being compared.

Calculus help needed?

First: Integrate MC.

The integral of 12x + 8 is ((12x^2)/2) +8x OR 6x^2 +8x

This is because the integral of x^n = (x^(n+1))/(n+1) and the integral of k = kx.

Then, just plug in the numbers from there :)

Calculus help needed!?

s (t) = [ 2t + 1 ]^(1/2)
s ` (t) = (1/2) (2) [ 2t + 1 ]^(-1/2)
s ` (t) = [ 2t + 1 ]^(-1/2)
s " (t) = (-1/2) [ 2t + 1 ]^(-3/2) [ 2 ]
s " (t) = - 1 / [ 2t + 1 ]^(3/2)
s " (4) = - 1 / [ 9 ]^(3/2) = - 1 / 27

Calculus Help (Need help with 1 problem)?

f '(x)= lim (h->0) f(x)
f '(x)= lim (h->0) (f(x+h)-f(x))/h
f '(x)= lim (h->0) ((4(x+h)^2-5(x+h))-(4x^2-5x.))/h
f '(x)= lim (h->0) ((4(h^2+2hx+x^2)-(5x+5h)-4x^2-5x.))/h
f '(x)= lim (h->0)(4h^2+8hx+4x^2-5x-5h-4x^2+5x.)/h
f '(x)= lim (h->0) (4h^2+8hx-5h)/h
factor out h so the h in the denominator cancels out
f '(x)= lim (h->0) (4h+8x-5)
sub h=0
f '(x)= 8x-5

Now just sub x=2 into the derivative to get the tangent at that point
f '(2)= 8(2)-5
f '(2)= 11

To find the equation of the line
m=(y2-y2)/(x2-x1)
11= (y-6)/(x-2)
y= 11x-16