91Ó°ÊÓ

A definite integral allows us to compute the area under a curve between two specific points on the x-axis. It is written in this format: \ \ \[ \ \ \int_{a}^{b} f(x) \, dx \ \ \] \ \ Here, \( a \) and \( b \) are the limits of integration, \( f(x) \) is the function being integrated, and '\( dx \)' indicates that we are integrating with respect to \( x \). \ \ Steps involved in calculating the definite integral: \ \

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• Determine the limits of integration.
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• Find the antiderivative of the function.
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• Evaluate the antiderivative at the upper limit and subtract the value of the antiderivative at the lower limit.
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\ \ In the given exercise, the definite integral is set up as: \ \ \ \ \[ \ \ \text{Area} = \ \ \int_{0}^{2} x^3 \, dx \ \ \] \ \ Evaluating this integral gives us the area under the curve from \( x = 0 \) to \( x = 2 \).

An antiderivative of a function \( f(x) \) is a function \( F(x) \) such that \( F'(x) = f(x) \). Essentially, it is the reverse of differentiation. For polynomial functions like \( y = x^3 \), we use the power rule for integration: \ \ \[ \ \ \ \ \int x^n \, dx = \ \ \frac{x^{n+1}}{n+1} + C \ \ \] \ \ Here, \( C \) is the constant of integration. In a definite integral, constants cancel out, so they are not included in our calculations. \ \ Steps to find the antiderivative: \ \

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• Identify the power of \( x \) in the function.
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• Apply the power rule.
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• Evaluate the new function, ignoring the constant \( C \).
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\ \ For the given function \( y = x^3 \), the antiderivative is: \ \ \[ \ \ \ \ \int x^3 \, dx = \frac{x^4}{4} \ \ \] \ \ We then use this antiderivative in the definite integral to find the area under the curve.

The area under a curve between two points on the x-axis can be found using the definite integral. This area is essentially the sum of infinitely small vertical slices from the curve down to the x-axis. \ \ To find this area, we follow these steps: \ \

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• Set up the definite integral with the given limits of integration.
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• Find the antiderivative of the function.
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• Evaluate the definite integral by finding the difference between the values of the antiderivative at the upper and lower limits.
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\ \ In our given example, the function is \( y = x^3 \) and the interval is [0,2]. We've already found the antiderivative to be \( \frac{x^4}{4} \). Now, we evaluate the definite integral: \ \ \[ \ \ \left[ \frac{x^4}{4} \right]_{0}^{2} = \ \ \left( \frac{2^4}{4} \right) - \left( \frac{0^4}{4} \right) = \ \ \frac{16}{4} - 0 = 4 \ \ \] \ \ Thus, the area under the curve from \( x = 0 \) to \( x = 2 \) is 4 square units.