Problem 23 - Algebra 2

Question

Find the area under each curve for the domain \(0 \leq x \leq 1\) $$ y=-(x-1)^{3}+3 $$

Step-by-Step Solution

Answer

The area under the curve \( y = -(x-1)^3 + 3 \) from \( x = 0 \) to \( x = 1 \) is \(\frac{5}{4}\).

Step 1: Understand the Integral

The formula to calculate the area under a curve represented by function \( f(x) \) between \( x = a \) and \( x = b \) is given by definite integral: \( \int_{a}^{b} f(x) dx \).

Step 2: Apply Corresponding Values

In our case, \( a = 0 \), \( b = 1 \), and the function \( f(x) = -(x-1)^3 + 3 \). So to find the area under the curve we have to calculate \( \int_{0}^{1} -(x-1)^3 + 3 dx \).

Step 3: Calculate the Integral

To calculate the integral we do small steps, let's change \( u = x - 1 \), then \( du = dx \). Change the limits accordingly, if \( x = 0 \), \( u = -1 \) and if \( x = 1 \), \( u = 0 \). So our integral turned into \( \int_{-1}^{0} -u^3 + 3 du \). Next, compute the anti-derivative of \( -u^3 + 3 \), which is \( -\frac{u^4}{4} + 3u \). Therefore, the area under the curve is given by: \( -\frac{u^4}{4} + 3u |_ {-1}^{0} \).

Step 4: Results

Plug in the values of the limits of the integral to get the area. \( [ -0^4/4 + 3*0 ] - [ -(-1)^4/4 + 3*(-1) ] = - 0 - ( -1/4 - 3 ) = - [ - \frac{5}{4} ] = \frac{5}{4} \).

Key Concepts

Definite Integral

A definite integral is a vital concept in calculus that allows for precise calculation of the area under a curve between two specified limits. It embodies the notion of accumulation and area computation with accuracy. Here, the function of interest is integrated between the bounds of the interval. The notation for a definite integral is generally expressed as \( \int_{a}^{b} f(x) \, dx \), which is read as "the integral of \( f(x) \) from \( a \) to \( b \)."

• \( a \) and \( b \) are the limits of integration.
• \( f(x) \) denotes the function you integrate.
• The result gives the net area between the curve and the x-axis within the given limits.

In our example, the function \( -(x-1)^3 + 3 \) is integrated from \( x = 0 \) to \( x = 1 \). Performing the integral yields the precise measurement of the area between the curve and the x-axis over this interval.

Area Under a Curve

Calculating the area under a curve is a fundamental task in calculus, revealing how much space is captured by the curve above or below the x-axis.
This area can represent various practical concepts, such as total distance traveled, volume of fluid, or any quantity that accumulates over time. Fundamentally, the area can be thought of as "adding up" infinitely small slices under the curve with widths approaching zero.

Example with Calculation

In the given function \( y = -(x-1)^3 + 3 \), we want the area from \( x = 0 \) to \( x = 1 \). This is efficiently done by computing the definite integral \( \int_{0}^{1} -(x-1)^3 + 3 \, dx \).

• Changing the integration variable simplifies the expression.
• The integration gives the aggregate area under the curve within the interval.

This calculation offers understanding into how accumulation occurs over the bounded range.

Integral Calculus

Integral calculus is a primary branch of calculus focused on the concept of integration. It provides tools for computing areas, volumes, and solving problems where an accumulation of quantities is necessary. Integration essentially reverses the process of differentiation, constructing antiderivatives of functions.

Understanding Integration Techniques

Various techniques and methods are applied in integral calculus to effectively solve integrals, such as:

• Substitution Method: Changing variables for simplifying integration, as shown when substituting \( u = x-1 \).
• Partial Fractions: Used for breaking intricate fractions for ease of integration.
• Integration by Parts: A method akin to the product rule of differentiation for product functions integration.

The solution of \( \int_{0}^{1} -(x-1)^3 + 3 \, dx \) shows practical application. By transforming the original variable \( x \) in the function, the integral becomes easier to manage. This demonstrates integral calculus’s importance in finding solutions to complex real-world problems effectively.