Final answer:
The antiderivative F(x) = x^2 + 5 is correctly identified for the integrand 2x. Using FTC^2, the integral from 0 to 1 of the function 2x dx is evaluated by subtracting the antiderivative at the lower bound from that at the upper bound, yielding the value of 1.
Step-by-step explanation:
Verification and Integral Evaluation
For the given integral \(\int_{0}^{1} 2x \, dx\), the function F(x) = x^2 + 5 is presented as an antiderivative of the integrand 2x. To verify this, we can differentiate F(x) to see if we obtain the original integrand. The derivative of F(x) with respect to x is 2x, which is indeed the integrand, confirming that F(x) is an antiderivative.
Using the Second Fundamental Theorem of Calculus (FTC^2), we can evaluate the integral by taking F(x) evaluated at the upper bound of the interval minus F(x) evaluated at the lower bound. Hence, the evaluation proceeds:
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- \(F(1) = 1^2 + 5 = 6\)
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- \(F(0) = 0^2 + 5 = 5\)
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- The integral \(\int_{0}^{1} 2x \, dx\) is then \(6 - 5 = 1\).
The answer is 1, which represents the area under the curve of the function from x=0 to x=1 shown in Figure 7.8, emphasizing the connection between integrals and areas under curves.