Question Statement

Evaluate the definite integral:

${\int }_{-1/2}^{3/2}(x-\cos \pi x)dx$

(Note: The original question contained $dv$ which has been corrected to $dx$ to match the solution context.)

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Background and Explanation

This problem involves computing a definite integral of a difference of functions. The key prerequisite is understanding the linearity of integration (allowing us to split the integral) and the Fundamental Theorem of Calculus (evaluating antiderivatives at the bounds).

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Solution

Step 1: Split the Integral

Using the linearity property of definite integrals, we can separate the integral of a difference into the difference of two integrals:

${\int }_{-1/2}^{3/2}(x-\cos \pi x)dx={\int }_{-1/2}^{3/2}xdx-{\int }_{-1/2}^{3/2}\cos \pi xdx$

Step 2: Find the Antiderivatives

For the first term, apply the power rule for integration: $\int xdx=\frac{x^2}{2}$

For the second term, use the substitution $u=\pi x$ (so $du=\pi dx$ or $dx=\frac{du}{\pi }$): $\int \cos \pi xdx=\frac{\sin \pi x}{\pi }$

Step 3: Apply the Fundamental Theorem of Calculus

Evaluate each antiderivative at the upper bound $x=\frac{3}{2}$ and lower bound $x=-\frac{1}{2}$:

First integral evaluation: ${\left[\frac{x^2}{2}\right]}_{-1/2}^{3/2}=\frac{1}{2}\left[{\left(\frac{3}{2}\right)}^2-{\left(-\frac{1}{2}\right)}^2\right]$

Calculate the squares: $=\frac{1}{2}\left[\frac{9}{4}-\frac{1}{4}\right]=\frac{1}{2}\left(\frac{8}{4}\right)=\frac{1}{2}(2)=1$

Second integral evaluation: ${\left[\frac{\sin \pi x}{\pi }\right]}_{-1/2}^{3/2}=\frac{1}{\pi }\left[\sin \left(\frac{3\pi }{2}\right)-\sin \left(-\frac{\pi }{2}\right)\right]$

Recall the sine values:

• $\sin \left(\frac{3\pi }{2}\right)=-1$
• $\sin \left(-\frac{\pi }{2}\right)=-1$ (since sine is odd: $\sin (-\theta )=-\sin (\theta )$ and $\sin (\frac{\pi }{2})=1$)

Substitute these values: $=\frac{1}{\pi }\left[-1-(-1)\right]=\frac{1}{\pi }(0)=0$

Alternative calculation shown in original: The evaluation can also be written as: $\frac{1}{\pi }\left[\sin \frac{3\pi }{2}+\sin \frac{\pi }{2}\right]=\frac{1}{\pi }[-1+1]=0$

Step 4: Combine Results

Subtract the second result from the first:

$I=1-0=1$

Therefore, the value of the definite integral is 1.

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Key Formulas or Methods Used

• Linearity of Integration: $\int [f(x)-g(x)]dx=\int f(x)dx-\int g(x)dx$
• Power Rule: $\int x^{n}dx=\frac{x^{n+1}}{n+1}$ (for $n\ne -1$)
• Integral of Cosine: $\int \cos (ax)dx=\frac{\sin (ax)}{a}$
• Fundamental Theorem of Calculus: ${\int }_a^{b}f(x)dx=F(b)-F(a)$ where $F^{\prime }(x)=f(x)$
• Trigonometric Values: $\sin (3\pi /2)=-1$, $\sin (\pi /2)=1$, and $\sin (-\theta )=-\sin (\theta )$

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Summary of Steps

1. Split the integral: $\int xdx-\int \cos \pi xdx$
2. Integrate each part: $\frac{x^2}{2}$ and $\frac{\sin \pi x}{\pi }$
3. Evaluate $\frac{x^2}{2}$ from $-1/2$ to $3/2$: $\frac{1}{2}(\frac{9}{4}-\frac{1}{4})=1$
4. Evaluate $\frac{\sin \pi x}{\pi }$ from $-1/2$ to $3/2$: $\frac{1}{\pi }(-1-(-1))=0$
5. Subtract to get final answer: $1-0=1$