Problem 36 - Algebra 2

Question

Evaluate the area under each curve for \(-1 \leq x \leq 2\) $$ g(x)=2^{x}+1 $$

Step-by-Step Solution

Answer

The area under the curve \( g(x)=2^{x}+1 \) from \(-1\) to \(2\) is \( \frac{7}{ln(2)} + 3 \).

Step 1: Setting Up the Integral

The first step involves setting up the proper integral to compute the area. The area under a curve \(f(x)\) from \(a\) to \(b\) is given by the definite integral from \(a\) to \(b\) of \(f(x)\). Therefore, to find the area under the curve \(g(x)=2^{x} + 1\) from \(-1\) to \(2\), the integral to evaluate is \(\int_{-1}^{2} (2^{x}+1) dx\).

Step 2: Evaluate the Integral

The next step is to evaluate the integral. It will be helpful to split the integral into two parts: \(\int_{-1}^{2} 2^{x} dx\) and \(\int_{-1}^{2} dx\). The first integral can be computed using the identity \(\int 2^{x} dx = \frac{2^{x}}{ln(2)} + C\). The second integral is quite straightforward: it is simply \(x\). Thus, the computation is now \[ \frac{2^{x}}{ln(2)} + x \] evaluated from \(-1\) to \(2\).

Step 3: Final Calculation

Finally, by substituting \(x = 2\) and \(x = -1\) into the equation from Step 2, we get \[ \frac{2^{2}}{ln(2)} + 2 - (\frac{2^{-1}}{ln(2)} - 1) \] = \[ \frac{4}{ln(2)} + 2 - \frac{1}{2*ln(2)} + 1\] = \[ \frac{7}{ln(2)} + 3 \] . This is the area under the curve.

Key Concepts

Area Under a Curve

When we talk about finding the "area under a curve," we are describing the process of determining the space between a given curve and the x-axis, over a specific interval. This is achieved using the concept of a definite integral. The definite integral of a function represents the accumulated area under its graph between two specified points on the x-axis.
For example, to find the area under the curve from \(x = a\) to \(x = b\) of a function \(f(x)\), we would calculate \(\int_{a}^{b} f(x) \, dx\).

• This integral sums the heights of infinitely small rectangles under the curve from \(x = a\) to \(x = b\).
• In our problem, this method was used on the function \(g(x) = 2^x + 1\) over the interval \([-1, 2]\).
• This involves splitting the integral if necessary and evaluating it part by part to find the total area.

Understanding this concept is crucial as it connects with many practical applications, such as calculating distances, probabilities, and even physics phenomena like work.

Exponential Functions

Exponential functions play a vital role in mathematics, characterized by a constant base raised to a variable exponent. A common example is \(f(x) = a^x\), where \(a\) is a base greater than zero, not equal to one, and \(x\) is the exponent that can be any real number.

• These functions are typically used to model growth processes, such as in population dynamics or radioactive decay.
• In the given exercise, our function \(g(x) = 2^x + 1\) includes an exponential component, \(2^x\).
• Exponential functions grow or decay at rates proportional to their current value, showcasing rapid increase or decrease.

In calculus, these functions often require specialized methods to integrate or differentiate, like using logarithmic identities. For instance, the integral \(\int 2^x \, dx\) applies the formula \(\frac{2^x}{\ln(2)} + C\), where \(C\) is the constant of integration. Understanding exponential functions is essential as they model real-world situations robustly linked to economics, biology, and engineering.

Integration by Parts

Integration by parts is a vital technique in calculus used to integrate products of functions. It's rooted in the product rule for differentiation and is particularly useful for handling situations where integration seems complex.
The formula for integration by parts is given by:\[ \int u \, dv = uv - \int v \, du \]Here, \(u\) and \(dv\) are chosen from the original integral, and the goal is to simplify the integration process.

• Identify \(u\) and \(dv\) to differentiate and integrate them respectively, resulting in a simpler integral.
• It helps transform difficult integrals into more manageable ones.
• This technique was crucial when breaking down and evaluating the given exponential and constant functions within the original exercise.

Though the specific original exercise uses a simpler form of integration, knowing integration by parts equips you with a versatile tool essential for more advanced calculus problems, especially those involving products of polynomials and logarithms.