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The concept of the integral arose from the need to calculate the area of a curved region. For example, calculating the area delimited by the region above the x-axis and below the graph of the function f(x) = x3 in a given limited interval is not such a simple task if we consider the knowledge acquired throughout basic education. But the definition of Definite Integral allows us to obtain this area precisely. Thus, given the function f(x) = x3, the area of the region below the graph of f, above the x-axis in the interval from 0 to 2 is equal to.

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Answer to a math question The concept of the integral arose from the need to calculate the area of a curved region. For example, calculating the area delimited by the region above the x-axis and below the graph of the function f(x) = x3 in a given limited interval is not such a simple task if we consider the knowledge acquired throughout basic education. But the definition of Definite Integral allows us to obtain this area precisely. Thus, given the function f(x) = x3, the area of the region below the graph of f, above the x-axis in the interval from 0 to 2 is equal to.

Expert avatar
Brice
4.8
109 Answers
1. \ \text{Evaluate the definite integral of the function:}
A = \int_{0}^{2} x^3 \, dx

2. \ \text{Find the antiderivative of} \ x^3:
\int x^3 \, dx = \frac{x^4}{4}

3. \ \text{Apply the limits of integration} \ (0 \ \text{to} \ 2):
A = \left[ \frac{x^4}{4} \right]_{0}^{2}

4. \ \text{Evaluate the antiderivative at the upper limit (2) and lower limit (0):}
A = \frac{2^4}{4} - \frac{0^4}{4}

5. \ \text{Simplify the expression:}
A = \frac{16}{4} - 0

6. \ \text{Finish the calculation:}
A = 4

\text{Therefore, the area is} \ 4

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