Problem 14
Question
Find the area of the region that lies under the graph of \(f\) over the given interval. $$f(x)=x+x^{2}, \quad 0 \leq x \leq 1$$
Step-by-Step Solution
Verified Answer
The area is \( \frac{5}{6} \).
1Step 1: Understand the Problem
We need to find the area under the curve of the function \( f(x) = x + x^2 \) between \( x = 0 \) and \( x = 1 \). This can be done by evaluating the definite integral of \( f(x) \) from 0 to 1.
2Step 2: Set Up the Integral
Set up the integral to find the area under the curve. We need to compute \( \int_{0}^{1} (x + x^2) \, dx \).
3Step 3: Integrate the Function
Find the indefinite integral of \( x + x^2 \). The integral of \( x \) is \( \frac{x^2}{2} \) and the integral of \( x^2 \) is \( \frac{x^3}{3} \). Thus, the indefinite integral of \( f(x) \) is \( \frac{x^2}{2} + \frac{x^3}{3} + C \).
4Step 4: Evaluate the Definite Integral
To find the definite integral from 0 to 1, we compute: \[ \int_{0}^{1} (x + x^2) \, dx = \left[ \frac{x^2}{2} + \frac{x^3}{3} \right]_{0}^{1} = \left( \frac{1^2}{2} + \frac{1^3}{3} \right) - \left( \frac{0^2}{2} + \frac{0^3}{3} \right). \] Calculate the expression to get: \[ \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}. \]
5Step 5: Conclusion
The area under the curve \( f(x) = x + x^2 \) from \( x = 0 \) to \( x = 1 \) is \( \frac{5}{6} \).
Key Concepts
Area Under a CurvePolynomial FunctionIntegration Techniques
Area Under a Curve
The term 'area under a curve' is quite self-explanatory. It refers to the total space or region that lies beneath a certain curve when plotted on a graph. In the context of calculus, this can be interpreted as finding the net value between the function and the x-axis, within a specified interval.
The area under a curve is often calculated using definite integrals. Importantly, this concept is pivotal in many real-life applications, like calculating distance from velocity-time graphs or finding total accumulation of quantities over time. By setting up a definite integral, you can effectively measure the space enclosed between the curve and the x-axis over the interval you're interested in.
For example, in this exercise, we found the area under the polynomial curve of the function \( f(x) = x + x^2 \) from \( x = 0 \) to \( x = 1 \). The area under the curve gives us the definite integral \( \int_{0}^{1} (x + x^2) \, dx \), resulting in an area of \( \frac{5}{6} \). This tells us the total "accumulated quantity" under that curve over the interval specified.
The area under a curve is often calculated using definite integrals. Importantly, this concept is pivotal in many real-life applications, like calculating distance from velocity-time graphs or finding total accumulation of quantities over time. By setting up a definite integral, you can effectively measure the space enclosed between the curve and the x-axis over the interval you're interested in.
For example, in this exercise, we found the area under the polynomial curve of the function \( f(x) = x + x^2 \) from \( x = 0 \) to \( x = 1 \). The area under the curve gives us the definite integral \( \int_{0}^{1} (x + x^2) \, dx \), resulting in an area of \( \frac{5}{6} \). This tells us the total "accumulated quantity" under that curve over the interval specified.
Polynomial Function
A polynomial function is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. In simpler terms, it consists of terms like \( x^2, x, \) or constants. These functions are fundamental, often serving as building blocks in algebra and calculus.
For this exercise, the polynomial function at hand is \( f(x) = x + x^2 \), which consists of two terms: a linear term \( x \) and a quadratic term \( x^2 \). Polynomials have various properties based on their degree, the highest power of the variable involved. Here, the highest degree is 2, making it a quadratic polynomial. Such functions generally form a parabolic curve when plotted on a graph.
Polynomial functions are continuous and smooth, making them predictable and easier to integrate. Calculating the area under a polynomial curve is straightforward when you apply integration techniques, as we've seen in this exercise using a definite integral to find the required area.
For this exercise, the polynomial function at hand is \( f(x) = x + x^2 \), which consists of two terms: a linear term \( x \) and a quadratic term \( x^2 \). Polynomials have various properties based on their degree, the highest power of the variable involved. Here, the highest degree is 2, making it a quadratic polynomial. Such functions generally form a parabolic curve when plotted on a graph.
Polynomial functions are continuous and smooth, making them predictable and easier to integrate. Calculating the area under a polynomial curve is straightforward when you apply integration techniques, as we've seen in this exercise using a definite integral to find the required area.
Integration Techniques
Integration techniques are essential tools in calculus used to find areas under curves and solve more complex problems.
The most common method is evaluating definite integrals, just like in this exercise. This involves finding the antiderivative or integral of the function followed by calculating the difference between the upper and lower bounds. Here, our function \( x + x^2 \) was integrated term by term.
Start by finding the indefinite integral:
For \( \int_{0}^{1} (x + x^2) \, dx \), this becomes \( \left[ \frac{x^2}{2} + \frac{x^3}{3} \right]_{0}^{1} \). You then compute and simplify: \( \frac{1^2}{2} + \frac{1^3}{3} \), finally yielding the solution of \( \frac{5}{6} \). Integration simplifies complex calculations into manageable computations, highlighting its importance in mathematics and applied sciences.
The most common method is evaluating definite integrals, just like in this exercise. This involves finding the antiderivative or integral of the function followed by calculating the difference between the upper and lower bounds. Here, our function \( x + x^2 \) was integrated term by term.
Start by finding the indefinite integral:
- The integral of \( x \) is \( \frac{x^2}{2} \)
- The integral of \( x^2 \) is \( \frac{x^3}{3} \)
For \( \int_{0}^{1} (x + x^2) \, dx \), this becomes \( \left[ \frac{x^2}{2} + \frac{x^3}{3} \right]_{0}^{1} \). You then compute and simplify: \( \frac{1^2}{2} + \frac{1^3}{3} \), finally yielding the solution of \( \frac{5}{6} \). Integration simplifies complex calculations into manageable computations, highlighting its importance in mathematics and applied sciences.
Other exercises in this chapter
Problem 13
Evaluate the limit, if it exists. $$\lim _{t \rightarrow-3} \frac{t^{2}-9}{2 t^{2}+7 t+3}$$
View solution Problem 13
Find the derivative of the function at the given number. $$f(x)=1-3 x^{2} \text { at } 2$$
View solution Problem 14
Evaluate the limit, if it exists. $$\lim _{h \rightarrow 0} \frac{\sqrt{1+h}-1}{h}$$
View solution Problem 14
Find the derivative of the function at the given number. $$f(x)=2-3 x+x^{2} \quad \text { at }-1$$
View solution