What is the result of evaluating the definite integral ∫ from 0 to 3 of (5) dx?

What is the result of evaluating the definite integral ∫ from 0 to 3 of (5) dx?

• A. 15
• B. 10
• C. 5
• D. 0

Answer: (A)

To evaluate the definite integral from 0 to 3 of the constant function 5, we can use the fundamental theorem of calculus. The integral of a constant \( c \) over an interval \([a, b]\) is given by the formula: \[ \int_a^b c \, dx = c(b - a) \] In this case, our constant \( c \) is 5, the lower limit \( a \) is 0, and the upper limit \( b \) is 3. Applying the formula: \[ \int_0^3 5 \, dx = 5(3 - 0) = 5 \times 3 = 15 \] Thus, the value of the definite integral is 15. This makes it clear that the integration of a constant over a defined range results in the constant multiplied by the width of the interval, which in this case is 3. Therefore, the correct answer is 15.

To evaluate the definite integral from 0 to 3 of the constant function 5, we can use the fundamental theorem of calculus. The integral of a constant ( c ) over an interval ([a, b]) is given by the formula:

[

\int_a^b c , dx = c(b - a)

]

In this case, our constant ( c ) is 5, the lower limit ( a ) is 0, and the upper limit ( b ) is 3. Applying the formula:

[

\int_0^3 5 , dx = 5(3 - 0) = 5 \times 3 = 15

]

Thus, the value of the definite integral is 15. This makes it clear that the integration of a constant over a defined range results in the constant multiplied by the width of the interval, which in this case is 3. Therefore, the correct answer is 15.