Solve the following integral. \int_{\frac{1}{2}}^{\frac{3}{2}}(- 2y + 4)dy

Question:

Solve the following integral.

{eq}\displaystyle \int_{\frac{1}{2}}^{\frac{3}{2}}(- 2y + 4)dy {/eq}

Definite Integral:

Integration is one of the major parts of the calculus. An integral with boundaries is called as a definite integral. To evaluate the integral first use the sum rule and then use the power rule to integrate. {eq}{\color{Blue}{\text{(The sum rule is} \displaystyle \int f(y) + g(y) dy = \int f(y) dy + \int g(y) dy)}} {/eq}

{eq}\begin{align*} \int_{\frac{1}{2}}^{\frac{3}{2}} (- 2y + 4)dy &= \int_{\frac{1}{2}}^{\frac{3}{2}} - 2y dy + \int_{\frac{1}{2}}^{\frac{3}{2}} 4 dy &{\color{Blue}{\text{(Apply the sum rule:} \displaystyle \int f(y) + g(y) dy = \int f(y) dy + \int g(y) dy)}} \\ &= - 2 \int_{\frac{1}{2}}^{\frac{3}{2}} y dy + 4 \int_{\frac{1}{2}}^{\frac{3}{2}} dy &\text{(Take the constant out)} \\ &= - 2 \left[\frac{y^2}{2}\right]_{\frac{1}{2}}^{\frac{3}{2}} + 4 \left[y\right]_{\frac{1}{2}}^{\frac{3}{2}} &{\color{Blue}{\text{(Apply the power rule:} \displaystyle \int y^n dx = \frac{y^{n + 1}}{n + 1}, where \ n \neq -1)}} \\ &= - 2 \left[\frac{9}{8} - \frac{1}{8}\right] + 4 \left[\frac{3}{2} - \frac{1}{2}\right] &\text{(Compute the boundaries)} \\ &= -2(1) + 4(1) \\ &= -2 + 4 \\ &= 2 \end{align*} {/eq}

{eq}\text{Therefore, the solution is} \boxed{ \ \displaystyle {\color{Red}{\int_{\frac{1}{2}}^{\frac{3}{2}}(- 2x + 4) dx = 2 }}} {/eq}