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Chapter 5: Problem 6

$$\text { Evaluate } \int_{0}^{2} 3 x^{2} d x \text { and } \int_{-2}^{2} 3 x^{2} d x$$

Short Answer

Expert verified
Question: Calculate the definite integrals of the function 3x^2 from 0 to 2 and from -2 to 2. Answer: The integral of 3x^2 from 0 to 2 is 8, and the integral of 3x^2 from -2 to 2 is 16.

Step by step solution

01

Apply the Power Rule for Integration

To integrate the function 3x^2, we can use the power rule for integration, which states that the integral of x^n with respect to x is x^(n+1)/(n+1) + C, where n is a constant and C is the integration constant. Let's apply the power rule to 3x^2: $$ \int 3x^2 dx = \frac{3}{3}x^{(2+1)} = x^3 + C $$
02

Evaluate the First Integral

Now we have the indefinite integral of 3x^2, which is x^3 + C. We want to find the integral of 3x^2 from 0 to 2, so we can use the Fundamental Theorem of Calculus, which says that to find the definite integral of a function over a certain interval, we can subtract the value of the antiderivative (indefinite integral) at the lower bound from the value of the antiderivative at the upper bound. In our case, the upper bound is 2 and the lower bound is 0. Let's find the definite integral: $$ \int_{0}^{2} 3x^2 dx = (2^3) - (0^3) = 8 - 0 = 8 $$
03

Evaluate the Second Integral

Using the same idea, we can now find the integral of 3x^2 from -2 to 2. Again, the indefinite integral is x^3 + C, so we can simply subtract the value of the antiderivative at the lower bound (-2) from the value of the antiderivative at the upper bound (2): $$ \int_{-2}^{2} 3x^2 dx = (2^3) - ((-2)^3) = 8 - (-8) = 8 + 8 = 16 $$ The final answer is that the integral of 3x^2 from 0 to 2 is 8, and the integral of 3x^2 from -2 to 2 is 16.

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