Question

Evaluate the following integral \(\int_1^2 x^3 \, dx\)

• A. \(\frac{4}{15}\)
• B. \(\frac{15}{4}\)
• C. \(\frac{7}{4}\)
• D. \(-\frac{4}{7}\)

Hint:

Always factor out constants before plugging in limits: \(\frac{1}{4}(2^4 - 1^4) = \frac{1}{4}(16 - 1) = \frac{15}{4}\) to minimize calculation errors.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The Fundamental Theorem of Calculus states that a definite integral can be evaluated by finding the antiderivative and subtracting its value at the lower limit from its value at the upper limit.
Key Formula or Approach:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1) \]
\[ \int_a^b f(x) \, dx = [F(x)]_a^b = F(b) - F(a) \]

Step 2: Detailed Explanation:
The antiderivative of \(x^3\) is:
\[ F(x) = \frac{x^4}{4} \]
Evaluating between the limits \(a = 1\) and \(b = 2\):
\[ \int_1^2 x^3 \, dx = \left[ \frac{x^4}{4} \right]_1^2 = \frac{2^4}{4} - \frac{1^4}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4} \]

Step 3: Final Answer:
Thus, the definite integral evaluates to \(\frac{15}{4}\), corresponding to option (B).