Problem 7
Question
Find the area under the given curve over the indicated interval. $$ y=x^{3} ; \quad[0,1] $$
Step-by-Step Solution
Verified Answer
The area under the curve from 0 to 1 is \( \frac{1}{4} \).
1Step 1: Interpret the Problem
We need to find the area under the curve defined by the function \( y = x^3 \) from \( x = 0 \) to \( x = 1 \). This requires us to compute a definite integral.
2Step 2: Set Up the Integral
The area under the curve can be found using the definite integral of the function over the given interval. We need to find \( \int_{0}^{1} x^3 \, dx \).
3Step 3: Apply Integration Rules
The integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. Here \( n = 3 \), so the antiderivative of \( x^3 \) is \( \frac{x^4}{4} \).
4Step 4: Evaluate the Definite Integral
Substitute the upper and lower limits into the antiderivative to find the definite integral: \( \left[ \frac{x^4}{4} \right]_{0}^{1} \). Calculate \( \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4} - 0 = \frac{1}{4} \).
5Step 5: Conclusion
The area under the curve \( y = x^3 \) from \( x = 0 \) to \( x = 1 \) is \( \frac{1}{4} \).
Key Concepts
Area under a curveIntegration rulesAntiderivativeEvaluating integrals
Area under a curve
The concept of finding the area under a curve is fundamental in calculus. Specifically, this area represents the accumulated "total" of all the y-values over a particular interval along the x-axis. This concept becomes particularly useful in a wide range of applications such as physics, engineering, and even economics, where it helps in understanding quantities like distance, volume, and cost.
To find the area under the curve, we use integrals. For the function in the exercise, \( y = x^3 \), over the interval [0,1], we aim to calculate this area using definite integrals. A definite integral takes into account the exact values at which we start and stop our measurement, which in this case, are \( 0 \) and \( 1 \). This precision allows us to accurately calculate the total area within those boundaries.
To find the area under the curve, we use integrals. For the function in the exercise, \( y = x^3 \), over the interval [0,1], we aim to calculate this area using definite integrals. A definite integral takes into account the exact values at which we start and stop our measurement, which in this case, are \( 0 \) and \( 1 \). This precision allows us to accurately calculate the total area within those boundaries.
Integration rules
Integration rules are guidelines that help in finding antiderivatives. These are formulas that assist in simplifying the process of integration, especially when dealing with polynomial functions like \( y = x^3 \).
The power rule is particularly important here: when you have \( x^n \), integrating it involves adding 1 to the power and dividing by this new power. Thus, the integration of \( x^3 \) results in \( \frac{x^4}{4} \).
The power rule is particularly important here: when you have \( x^n \), integrating it involves adding 1 to the power and dividing by this new power. Thus, the integration of \( x^3 \) results in \( \frac{x^4}{4} \).
- Always remember, these rules generalize to any polynomial, handle each one according to its power.
- Don’t forget the constant of integration, \( C \), when finding indefinite integrals; however, it vanishes in definite integrals since we evaluate between two specific limits.
Antiderivative
The antiderivative is essentially the reverse of differentiation. In simpler terms, if differentiation is about finding the rate of change, antiderivatives recover the original function from its derivative. In calculus, if you start from a rate of growth, the antiderivative gives you the accumulated total or quantity.
For instance, given the function \( y = x^3 \) and asked to find its antiderivative, we apply the power rule of integration. Doing so yields the result \( \frac{x^4}{4} \). This new expression lets us understand how \( y = x^3 \) accumulates or grows from base \( x^3 \) values.
Antiderivatives play a crucial role when evaluating definite integrals to get a precise measure of accumulated quantities within specified limits.
For instance, given the function \( y = x^3 \) and asked to find its antiderivative, we apply the power rule of integration. Doing so yields the result \( \frac{x^4}{4} \). This new expression lets us understand how \( y = x^3 \) accumulates or grows from base \( x^3 \) values.
Antiderivatives play a crucial role when evaluating definite integrals to get a precise measure of accumulated quantities within specified limits.
Evaluating integrals
Evaluating definites integrals is the final step in finding the exact area under a curve for a given interval. This process essentially transitions from having an antiderivative to determining an actual numerical value.
In the problem \( \int_{0}^{1} x^3 \, dx \), we first derive the antiderivative, \( \frac{x^4}{4} \). Then, we substitute the upper limit (\( x=1 \)) and lower limit (\( x=0 \)) into this expression:
Thus, the evaluated integral gives us \( \frac{1}{4} \), which is the area under the curve from \( x = 0 \) to \( x = 1 \). This method solidifies the link between abstract computation and tangible area measurement.
In the problem \( \int_{0}^{1} x^3 \, dx \), we first derive the antiderivative, \( \frac{x^4}{4} \). Then, we substitute the upper limit (\( x=1 \)) and lower limit (\( x=0 \)) into this expression:
- Substitute 1: \( \frac{1^4}{4} = \frac{1}{4} \)
- Substitute 0: \( \frac{0^4}{4} = 0 \)
Thus, the evaluated integral gives us \( \frac{1}{4} \), which is the area under the curve from \( x = 0 \) to \( x = 1 \). This method solidifies the link between abstract computation and tangible area measurement.
Other exercises in this chapter
Problem 7
Evaluate using integration by parts or substitution. (Assume \(u>0\) in \(\ln\) u. Check by differentiating. $$ \int x e^{-2 x} d x $$
View solution Problem 7
Find each integral. $$ \int\left(x^{2}+x-1\right) d x $$
View solution Problem 7
Find the area represented by each definite integral. $$ \int_{0}^{4}|x-3| d x $$
View solution Problem 8
Evaluate. (Be sure to check by differentiating!) $$ \int \frac{5}{5 x+7} d x $$
View solution