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Chapter 5: Problem 42

Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{0}^{\pi}(1-\sin x) d x$$

Short Answer

Expert verified
Question: Evaluate the integral of the function $$f(x) = 1 - \sin x$$ in the interval $$[0, \pi]$$. Answer: The value of the integral is $$\pi + 2$$.

Step by step solution

01

Find the antiderivative of the function

To find the antiderivative of $$f(x) = 1 - \sin x$$, we will find the antiderivative of each term and combine them: $$\int(1-\sin x) d x = \int 1 dx - \int \sin x dx$$ Now, we will find the antiderivative of each term separately: $$\int 1 dx = x + C_1$$ $$\int \sin x dx= -\cos x + C_2$$ Therefore, the antiderivative of $$f(x)$$ is: $$F(x) = x -\cos x + C$$
02

Apply the Fundamental Theorem of Calculus

To evaluate the integral using the Fundamental Theorem of Calculus, we will plug in the values $$a = 0$$ and $$b = \pi$$ into the antiderivative $$F(x)$$ and find the difference between them: $$\int_{0}^{\pi}(1-\sin x)d x = F(\pi) - F(0)$$ Now we will substitute the values and compute the integral: $$F(\pi) = \pi - \cos(\pi) = \pi - (-1) = \pi + 1$$ $$F(0) = 0 - \cos(0) = 0 - 1 = -1$$ Finally, we will find the difference: $$\int_{0}^{\pi}(1-\sin x) d x = F(\pi) - F(0) = (\pi + 1)-(-1) = \pi + 2$$ Thus, the value of the integral is $$\pi + 2$$.

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