Problem 9
Question
Compute the area of the region bounded by the graph of \(y=t^{2},\) the \(t\) -axis, and the lines \(t=1\) and \(t=2\).
Step-by-Step Solution
Verified Answer
The area is \(\frac{7}{3}\).
1Step 1: Set Up the Integral
To find the area under the curve of the function \(y = t^2\) from \(t = 1\) to \(t = 2\), we need to set up the definite integral. The area \(A\) is given by the integral: \[A = \int_{1}^{2} t^2 \, dt\]
2Step 2: Integrate the Function
Find the antiderivative of \(t^2\), which is \(\frac{t^3}{3}\). We use this to evaluate the definite integral:\[A = \left[ \frac{t^3}{3} \right]_{1}^{2}\]
3Step 3: Evaluate the Definite Integral
Substitute the upper limit \(t = 2\) into the antiderivative:\[\frac{(2)^3}{3} = \frac{8}{3}\] Substitute the lower limit \(t = 1\) into the antiderivative:\[\frac{(1)^3}{3} = \frac{1}{3}\]Subtract the value of the lower limit from the upper limit:\[A = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}\]
4Step 4: Simplify the Result
The result \(\frac{7}{3}\) is already in its simplest form, representing the area of the region bounded by the curve \(y = t^2\) between \(t = 1\) and \(t = 2\).
Key Concepts
Area Under a CurveAntiderivativeEvaluation of Limits
Area Under a Curve
Finding the "area under a curve" is a way to determine how much space is enclosed between the curve of a function and the horizontal axis over an interval. In the context of definite integrals, this area is mathematically represented by the definite integral of a function over a specific range. For example, to find the area under the curve for the function \( y = t^2 \) from \( t = 1 \) to \( t = 2 \), we calculate the definite integral of \( t^2 \) with respect to \( t \).
This process involves setting up the integral with the appropriate boundaries:
By integrating, we summarize the infinite number of infinitely small "pieces of area" beneath the curve, cumulatively resulting in the total area from \( t = 1 \) to \( t = 2 \).
Hence, the integral will give us the complete area under the curve, representing the bounded region we aim to find.
This process involves setting up the integral with the appropriate boundaries:
- The lower limit of integration corresponds to the starting point, here \( t = 1 \).
- The upper limit of integration is the endpoint, here \( t = 2 \).
By integrating, we summarize the infinite number of infinitely small "pieces of area" beneath the curve, cumulatively resulting in the total area from \( t = 1 \) to \( t = 2 \).
Hence, the integral will give us the complete area under the curve, representing the bounded region we aim to find.
Antiderivative
An "antiderivative" is essentially the opposite of taking a derivative. It is a function that undoes differentiation, essentially finding the original function given its derivative. In calculus, when faced with finding the area under a curve, we often need the antiderivative to perform the integration process. Suppose we have the function \( y = t^2 \). The goal is to use its antiderivative to solve the definite integral.
The antiderivative of \( t^2 \) is \( \frac{t^3}{3} \), because when you take the derivative of \( \frac{t^3}{3} \), you get back to \( t^2 \).
The antiderivative is crucial when you evaluate a definite integral, as it allows us to find the area under the curve by applying the Fundamental Theorem of Calculus:
The antiderivative of \( t^2 \) is \( \frac{t^3}{3} \), because when you take the derivative of \( \frac{t^3}{3} \), you get back to \( t^2 \).
The antiderivative is crucial when you evaluate a definite integral, as it allows us to find the area under the curve by applying the Fundamental Theorem of Calculus:
- Compute the antiderivative.
- Evaluate it at the upper limit of integration.
- Subtract the result of evaluating it at the lower limit.
Evaluation of Limits
"Evaluation of limits" in the context of definite integrals refers to calculating the finite area under a curve between two bounds. After finding the antiderivative of a function, the next step involves substituting the upper and lower bounds to find specific values that define this area.
Taking the function \( y = t^2 \) as an example, once the antiderivative \( \frac{t^3}{3} \) is found, the evaluation proceeds by substituting the bounds:\[ \left[ \frac{t^3}{3} \right]_1^2 \]
The upper limit \( t = 2 \) is substituted into the antiderivative to find \( \frac{8}{3} \), and the lower limit \( t = 1 \) is substituted to obtain \( \frac{1}{3} \).
The evaluation of limits translates to finding the difference between these two values, enabling us to effectively calculate the definite integral:
Taking the function \( y = t^2 \) as an example, once the antiderivative \( \frac{t^3}{3} \) is found, the evaluation proceeds by substituting the bounds:\[ \left[ \frac{t^3}{3} \right]_1^2 \]
The upper limit \( t = 2 \) is substituted into the antiderivative to find \( \frac{8}{3} \), and the lower limit \( t = 1 \) is substituted to obtain \( \frac{1}{3} \).
The evaluation of limits translates to finding the difference between these two values, enabling us to effectively calculate the definite integral:
- Find \( \frac{8}{3} \): this is the area from 0 to \( t = 2 \).
- Find \( \frac{1}{3} \): this is the area from 0 to \( t = 1 \).
- Subtract \( \frac{1}{3} \) from \( \frac{8}{3} \): this isolates the area purely between \( t = 1 \) to \( t = 2 \).
Other exercises in this chapter
Problem 9
Suppose \(f,\) is a continuous defined on an interval \([a, b]\) and \((v, f(v))\) is a high point of \(f\) on \([a, b]\) (meaning that \(v\) is in \([a, b]\) a
View solution Problem 9
The exact area of the region bounded by the graphs of $$y=\sqrt{t} \quad y=0 \quad \text { and } \quad t=5$$ is \(\frac{2}{3} 5^{3 / 2} \doteq 7.453560\). Compu
View solution Problem 10
The exact area of the region bounded by the graphs of $$y=\sqrt{t} \quad y=0 \quad \text { and } \quad t=5 $$ is \(\frac{2}{3} 5^{3 / 2} \doteq 7.453560\). Comp
View solution Problem 11
Write an integral that is the area of the region bounded by the graphs of a. \(y=t^{2}-t \quad\) and \(\quad y=0, \quad t=1, \quad t=2\). b. \(y=t^{2}, \quad\)
View solution