Calc 1 Final Questions

Question 1

What is the Trapezoid Rule and how is it applied if n = 4 ?

Answer 1

Trapezoid Rule is a means of summation to estimate the area under a curve using the equation:

For n = 4

[ ƒ(x₀) + 2ƒ(x₁) + 2ƒ(x₂) + 2ƒ(x₃) + ƒ(x₄) ]

Question 2

How do is the equation that best represents the shaded region found ?

Answer 2

-Take integral of the topmost function subtracted by the bottomost function.
-For the bounds, set the equations equal to each other.

∫₀³ [(-x² + 4x) - (x² - 2x)]dx

Question 3

How can this function be converted into an integral ?

Answer 3

Start by noticing the D/n in the function as the equation for Δx = (b-a)/n. We can assume our lower bound is 0 and upper bound is D

Also recognized the constant at the beginning of the function c.

Lastly note that the xₖ** which falls into ƒ(xₖ) is x*ₖ = (a +Δx(k))

Question 4

What is the position function s(t) when v(t) = t² + 2t
and the initial position is 4 meters

Answer 4

Since 𝑠(𝑡) = 4 and 𝑠(𝑡) = ∫𝑣(𝑡)𝑑𝑡

𝑠(𝑡) = (1/3)t³ + t² + 4

Question 5

Use differentials to approximate the change in the volume of a sphere
when its radius changes from 𝑟 = 3 centimeters to 𝑟 = 3.05 centimeters.
Recall: Volume of a sphere is given by the formula 𝑉 = (4/3)π𝑟³

Answer 5

Since we know the typical equation for a differential is dy = ƒ(x)dx where dx represents the change in x we can start by finding the differential equation with respect to 𝑟.

𝑑v = 4π𝑟² 𝑑𝑟

Since we know that 𝑑𝑟 = 𝑟₁ - 𝑟₀ = [3.05 -3]cm = .05cm

Thus our function is 𝑑v = 4π(3cm)² (.05cm) = (9/5)πcm³

Question 6

Find the AntiDerivative of:
∫ x∛(x+2)𝑑𝑥

Answer 6

Start by using substitution to where u = x + 2 and du = 1dx

then solve for x to help eliminate the x within the function: x = u - 2.

Now we have: ∫ (u - 2)∛(u)du → ∫ ∛(u⁴) - 2∛(u)du

With this easier equation, we can now solve the integral:

(3/7)∛(u⁷) - (3/2)∛(u⁴) → (3/7)∛((x + 2)⁷) - (3/2)∛((x + 2)⁴) + c

Question 7

Find the definite integral of:

∫₀ ˡⁿ⁽⁸⁾ eˣ𝑑𝑥

Answer 7

Note that ∫eˣ𝑑𝑥 = eˣ

∫₀ ˡⁿ⁽⁸⁾ eˣ𝑑𝑥 = [eˡⁿ⁽⁸⁾ - e⁰] = [8 - 1] = 7

Question 8

Simplify the expression:

𝑑/𝑑𝑥 ∫ₓ² (t³ - 4t²)𝑑𝑡

Answer 8

Due to the Fundamental Theorem of Calculus we can sub 𝑥 into the values of t. However since 𝑥 is in the lower bound we will negate the function:

𝑑/𝑑𝑥 ∫ₓ² (t³ - 4t²)𝑑𝑡 = - (𝑥³ - 4𝑥²) → -𝑥³ + 4𝑥²

Question 9

Is this Limit in Indeterminate Form and if so what type ?

Limit x → - π/2⁺ of tanx ⋅ (x + π/2)

Answer 9

∞ ⋅ 0

Question 10

How to Calculate Midpoint Riemann Sum over interval [-1, 5] if:

ƒ(x) = x³ + 1 and n = 3

Answer 10

First define our Δx as (5 + 1)/3 = 2

Then define our subintervals which are -1 + Δx until we reach 5.

So x = [_-1_1_35] then we take the midpoint of each of these values:

x = [ _0_24 ] then we will perform Riemann’s sum with our new midpoint values:

Δx[ ƒ(x₁) + ƒ(x₂) + ƒ(x₃) ].