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Use limits to evaluate the integral.

[tex]\[ \int_0^2 7x^3 \, dx \][/tex]

A. 14
B. 2
C. 28
D. 112


Sagot :

To evaluate the integral [tex]\(\int_0^2 7 x^3 \, dx\)[/tex] using limits, we follow these steps:

1. Identify the integral to solve:
[tex]\[ \int_0^2 7 x^3 \, dx \][/tex]

2. Find the antiderivative of the integrand:
The integrand is [tex]\(7 x^3\)[/tex]. To find the antiderivative, we use the power rule for integration, [tex]\(\int x^n \, dx = \frac{x^{n+1}}{n+1}\)[/tex], and multiply by the constant coefficient.
[tex]\[ \int 7 x^3 \, dx = 7 \int x^3 \, dx = 7 \left( \frac{x^{3+1}}{3+1} \right) = 7 \left( \frac{x^4}{4} \right) = \frac{7}{4} x^4 \][/tex]

3. Evaluate the definite integral at the limits from 0 to 2:
We need to compute the difference between the antiderivative evaluated at the upper limit (2) and the lower limit (0).
[tex]\[ \left[ \frac{7}{4} x^4 \right]_0^2 = \left( \frac{7}{4} (2^4) \right) - \left( \frac{7}{4} (0^4) \right) \][/tex]

4. Calculate the values at the upper and lower limits:
First, evaluate at [tex]\(x = 2\)[/tex]:
[tex]\[ \frac{7}{4} (2^4) = \frac{7}{4} (16) = \frac{7 \times 16}{4} = \frac{112}{4} = 28 \][/tex]
Then, evaluate at [tex]\(x = 0\)[/tex]:
[tex]\[ \frac{7}{4} (0^4) = \frac{7}{4} (0) = 0 \][/tex]

5. Subtract the lower limit evaluation from the upper limit evaluation:
[tex]\[ 28 - 0 = 28 \][/tex]

Therefore, the value of the definite integral [tex]\(\int_0^2 7 x^3 \, dx\)[/tex] is [tex]\(\boxed{28}\)[/tex].
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