Given integral subscript short dash 4 end subscript superscript 11 f left parenthesis x right parenthesis d x equals 34 and integral subscript short dash 4 end subscript superscript 11 g left parenthesis x right parenthesis d x equals short dash 5, find the following:

integral subscript 11 superscript short dash 4 end superscript left square bracket 4 f left parenthesis x right parenthesis minus 3 g left parenthesis x right parenthesis right square bracket d x

To solve the given integral:

\[ \int_{11}^{-4} [4f(x) - 3g(x)] \, dx, \]

we need to make use of the properties of integrals and the given values:

\[ \int_{-4}^{11} f(x) \, dx = 34 \]
\[ \int_{-4}^{11} g(x) \, dx = -5 \]

First, let's apply the property of integral limits. Changing the limits of integration will change the sign of the integral:

\[ \int_{a}^{b} h(x) \, dx = -\int_{b}^{a} h(x) \, dx \]

Thus,

\[ \int_{11}^{-4} f(x) \, dx = -\int_{-4}^{11} f(x) \, dx \]
\[ \int_{11}^{-4} f(x) \, dx = -34 \]

Similarly,

\[ \int_{11}^{-4} g(x) \, dx = -\int_{-4}^{11} g(x) \, dx \]
\[ \int_{11}^{-4} g(x) \, dx = 5 \]

Now we can use the linearity of the integral to split it into parts:

\[ \int_{11}^{-4} [4f(x) - 3g(x)] \, dx = 4 \int_{11}^{-4} f(x) \, dx - 3 \int_{11}^{-4} g(x) \, dx \]

Substituting the values we have calculated:

\[ 4 \int_{11}^{-4} f(x) \, dx = 4(-34) = -136 \]
\[ -3 \int_{11}^{-4} g(x) \, dx = -3(5) = -15 \]

Adding these results together:

\[ \int_{11}^{-4} [4f(x) - 3g(x)] \, dx = -136 - 15 = -151 \]

Therefore, the value of the integral is:

\[ \int_{11}^{-4} [4f(x) - 3g(x)] \, dx = -151 \]

Would you like any further details or have any questions?

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1. How do you find the derivative of a product of two functions?
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**Tip:** When solving integrals, always consider whether you can simplify the problem using symmetry, linearity, or known values of integrals over the same interval.

Learn how to calculate the integral of [4f(x) - 3g(x)] from 11 to -4 using the properties of integrals and given values. This problem involves applying linearity of integrals and changing limits to solve the integral effectively.

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