Evaluate the following integral: ∫ 0 3 4 d x \int_0^3\!4\,dx ∫ 0 3 4 d x

in this problem, we are asked to evaluate the following indefinite integral. Now we've learned from the fundamental theorem of calculus part two that we can do these by combining the idea of taking an antiderivative and just evaluating these bounds together. So what we can do is say, alright. So we have this four here that we're integrating from zero to three with respect to x. Now we know that the antiderivative of a constant with respect to x is just gonna be that constant times x. So the antiderivative here is just gonna be four x. Then I need to go ahead and bound this from zero to three. So we've gone ahead and applied this rule. Now what I need to do is plug in the high bound and the low bound and subtract. So we'll start with the high bound. So the high bound is gonna give us four times three. Then we're going to have minus, and I'll plug in the low bound, which is gonna be four times zero. Notice I'm just taking this x here, and they're replacing it with three and with zero. Now also something I wanna mention here is notice I did not add a plus c to this because we don't need to do that when we're dealing with definite integrals. So we have four times three minus four times zero. Four times three is 12, then we have that minus zero, and the whole thing is just gonna come out to 12, meaning 12 is the solution to our problem. So that's how we can find this definite integral. Hope you found this video helpful.

8. Definite Integrals

Fundamental Theorem of Calculus

Business Calculus

8. Definite Integrals

Fundamental Theorem of Calculus

Struggling with Business Calculus?

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Multiple Choice

Evaluate the following integral:
$\int_0^3\!4\,dx$

Verified step by step guidance

Recognize that the integral ∫₀³ 4 dx represents the area under the constant function f(x) = 4 from x = 0 to x = 3.

Since the function f(x) = 4 is constant, the integral simplifies to multiplying the constant value by the length of the interval: ∫₀³ 4 dx = 4 × (3 - 0).

Calculate the length of the interval by subtracting the lower limit of integration (0) from the upper limit of integration (3).

Multiply the constant value (4) by the length of the interval (3 - 0) to find the result of the integral.

The final result represents the total area under the curve of f(x) = 4 from x = 0 to x = 3.

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Evaluate the following integral:∫034dx\int_0^3\!4\,dx∫034dx

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Fundamental Theorem of Calculus practice set

Evaluate the following integral:
$\int_0^3\!4\,dx$