Math Problem Statement

1. Diberikan kurva y = x^3 - 3x^2 + 4 dan sumbu-x. Hitunglah luas daerah yang dibatasi oleh kurva tersebut dan sumbu-x pada interval x ∈ [1, 4]. 2. Sebuah benda bergerak sepanjang garis lurus dengan kecepatan v(t) = t^2 - 4t + 3 dalam meter per detik, dengan t dalam detik. a. Tentukan perpindahan total benda dalam interval t ∈ [0, 5]. b. Hitung jarak total yang ditempuh benda dalam interval t ∈ [0, 5]. 3. Diberikan fungsi komposit y = ln(√(x^2 + e^(sin(3x)))) · cos^4(1/x). Hitunglah turunan pertama dy/dx dengan menggunakan aturan rantai, aturan hasil kali, dan aturan turunan fungsi trigonometri.

Solution

Let's solve the given problems step by step.


Soal 1

Given: The curve y=x33x2+4y = x^3 - 3x^2 + 4 and the x-axis, over the interval x[1,4]x \in [1, 4].
Find: The area between the curve and the x-axis.

Solution

The formula to compute the area between a curve y=f(x)y = f(x) and the x-axis is:

Area=abf(x)dx\text{Area} = \int_{a}^{b} |f(x)| \, dx

Here, f(x)=x33x2+4f(x) = x^3 - 3x^2 + 4, and the interval is [1,4][1, 4]. First, we check where the curve crosses the x-axis by solving f(x)=0f(x) = 0:

x33x2+4=0x^3 - 3x^2 + 4 = 0

This can be solved to determine points of sign changes. Then, integrate separately where f(x)f(x) is positive or negative.


Soal 2

Given: v(t)=t24t+3v(t) = t^2 - 4t + 3 is the velocity of a particle over t[0,5]t \in [0, 5].

Part (a): Total Displacement

The total displacement is the integral of the velocity function:

Displacement=05v(t)dt\text{Displacement} = \int_0^5 v(t) \, dt

Substitute v(t)=t24t+3v(t) = t^2 - 4t + 3, and evaluate.

Part (b): Total Distance

To find the total distance, we integrate the absolute value of the velocity:

Distance=05v(t)dt\text{Distance} = \int_0^5 |v(t)| \, dt

Identify when v(t)=0v(t) = 0 to check where the velocity changes sign. Solve:

t24t+3=0    (t3)(t1)=0    t=1,t=3t^2 - 4t + 3 = 0 \implies (t-3)(t-1) = 0 \implies t = 1, t = 3

Thus, split the integral into intervals [0,1][0,1], [1,3][1,3], and [3,5][3,5], adjusting for the sign of v(t)v(t) in each interval.


Soal 3

Given: y=ln(x2+esin(3x))cos4(1x)y = \ln\left( \sqrt{x^2 + e^{\sin(3x)}} \right) \cdot \cos^4\left( \frac{1}{x} \right).
Find: dydx\frac{dy}{dx}.

Solution

To differentiate yy, we use:

  1. Product Rule: (uv)=uv+uv(u \cdot v)' = u'v + uv'
  2. Chain Rule for composite functions.
  3. Trigonometric Derivatives: Differentiating cosn(u)\cos^n(u) and related functions.

Let u=ln(x2+esin(3x))u = \ln\left( \sqrt{x^2 + e^{\sin(3x)}} \right) and v=cos4(1x)v = \cos^4\left( \frac{1}{x} \right).

  • Differentiate uu using the chain rule:
    u=ln(x2+esin(3x))u = \ln\left( \sqrt{x^2 + e^{\sin(3x)}} \right)

  • Differentiate vv using the chain and power rule.

Combine both results using the product rule to find dydx\frac{dy}{dx}.


Would you like step-by-step calculations for any of these problems? Let me know!


5 Related Questions:

  1. How do you calculate areas under a curve if the function crosses the x-axis?
  2. What is the difference between displacement and distance in kinematics?
  3. How do you solve integrals involving absolute values?
  4. How do you use the product rule for differentiation?
  5. How do you apply the chain rule to logarithmic and trigonometric functions?

Tip: When solving for total distance, always split the integral at points where the velocity changes sign to ensure you account for magnitude correctly.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Integration (Definite Integrals)
Displacement and Total Distance in Kinematics
Differentiation: Chain Rule, Product Rule, Trigonometric Derivatives

Formulas

Area under a curve: A = ∫[a,b] |f(x)| dx
Displacement: ∫ v(t) dt
Total distance: ∫ |v(t)| dt
Chain Rule: d/dx [f(g(x))] = f'(g(x))·g'(x)
Product Rule: d/dx [u·v] = u'v + uv'

Theorems

Fundamental Theorem of Calculus
Rules of Differentiation
Absolute Value in Integrals

Suitable Grade Level

Undergraduate Mathematics or Advanced High School Calculus