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If integral_{-60}^{-30} f (x) dx = 15 and integral_{-60}^{-39} g (x) dx = 28 and...

Question:

If {eq}\displaystyle \int_{-60}^{-39} f (x)\ dx = 15 {/eq} and {eq}\displaystyle \int_{-60}^{-39} g (x)\ dx = 28 {/eq} and {eq}\displaystyle \int_{-60}^{-39} h (x)\ dx = 25 {/eq}, what does the following integral equal?

{eq}\displaystyle \int_{-60}^{-39} [5 f(x) + 4 g(x) - h (x)]\ dx {/eq}.

Techniques for Definite Integral:

If there are some individual definite integral of functions and we need to find a combined integral of these functions, then we apply the sum and subtract rule of definite integral to write the combination as the sum of the individual integrals.

  • {eq}\int_{a}^{b}(p(x)+q(x)-r(x))\ dx=\int_{a}^{b}p(x)\ dx+\int_{a}^{b}q(x)\ dx-\int_{a}^{b}r(x)\ dx {/eq}

Answer and Explanation:

Given data:

{eq}\displaystyle \int_{-60}^{-39} f (x)\ dx = 15\\ \displaystyle \int_{-60}^{-39} g (x)\ dx = 28\\ \displaystyle \int_{-60}^{-39} h (x)\ dx = 25\\ \displaystyle \int_{-60}^{-39} [5 f(x) + 4 g(x) - h (x)]\ dx=? {/eq}

Applying the sum and subtract rule of integration, we get:

{eq}\displaystyle \int_{-60}^{-39} [5 f(x) + 4 g(x) - h (x)]\ dx=\int_{-60}^{-39} 5 f(x)\ dx+\int_{-60}^{-39} 4 g(x)\ dx-\int_{-60}^{-39}h(x)\ dx {/eq}

Take out the constant and substitute the given values of the definite integral in the above expression.

{eq}\begin{align*} \displaystyle \int_{-60}^{-39} 5 f(x)\ dx+\int_{-60}^{-39} 4 g(x)\ dx-\int_{-60}^{-39}h(x)\ dx&=5\int_{-60}^{-39} f(x)\ dx+4\int_{-60}^{-39} g(x)\ dx-\int_{-60}^{-39}h(x)\ dx\\ &=5(15)+4(28)-25\\ &=75+112-25\\ &=187-25\\ &=162 \end{align*} {/eq}


Learn more about this topic:

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Evaluating Definite Integrals Using the Fundamental Theorem

from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2
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