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Chapter 5: Problem 10

$$\text { Explain why } \int_{a}^{b} f^{\prime}(x) d x=f(b)-f(a)$$

Short Answer

Expert verified
Answer: The integral of the derivative of a function from a to b equals the function's values at b minus its values at a because of the Fundamental Theorem of Calculus, which connects differentiation and integration. The theorem states that for a continuous function on the interval [a, b], the integral of its derivative is equal to the difference in the function's value at those points: $\int_{a}^{b} f^{\prime}(x) d x=f(b)-f(a)$.

Step by step solution

01

Recall the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that: if a function F is continuous on the interval [a, b] and F'(x) = f(x) for all x belonging to [a, b], then $$\int_{a}^{b} f(x) d x = F(b) - F(a)$$ In the given question, we are asked to explain why the integral of f'(x) is equal to the difference between f(b) and f(a). We can apply the Fundamental Theorem of Calculus here.
02

Apply the Fundamental Theorem of Calculus

Given that f'(x) is the derivative of function f(x), we can treat f'(x) as g(x) for simplicity, and let G(x) be its antiderivative: $$G'(x) = g(x) = f'(x)$$ Now we can apply the Fundamental Theorem of Calculus: $$\int_{a}^{b} f'(x) d x = \int_{a}^{b} g(x) d x = G(b) - G(a)$$
03

Use the antiderivative of f'(x)

Since f'(x) is the derivative of f(x) and G'(x) = f'(x), it means that G(x) is an antiderivative of f'(x). Thus, G(x) is equal to f(x) plus some constant, C: $$G(x) = f(x) + C$$ Now, we can substitute this equation with G(b) and G(a): $$\int_{a}^{b} f'(x) d x = f(b) + C - f(a) - C$$ The constants, C, will cancel each other out: $$\int_{a}^{b} f'(x) d x = f(b) - f(a)$$
04

Conclusion

Based on the Fundamental Theorem of Calculus and using the antiderivative of f'(x), we have shown that the integral of a function's derivative from a to b is equal to the difference in the function's value at those points: $$\int_{a}^{b} f^{\prime}(x) d x=f(b)-f(a)$$

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Most popular questions from this chapter

Consider the function \(f(x)=x^{2}-4 x\) a. Graph \(f\) on the interval \(x \geq 0\) b. For what value of \(b>0\) is \(\int_{0}^{b} f(x) d x=0 ?\) c. In general, for the function \(f(x)=x^{2}-a x,\) where \(a>0,\) for what value of \(b>0\) (as a function of \(a\) ) is \(\int_{0}^{b} f(x) d x=0 ?\)

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