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Using linearity and bounds-reversal to evaluate a combined integral

Suppose ∫15f(x) dx=8\int_1^5 f(x)\,dx = 8 and ∫15g(x) dx=3\int_1^5 g(x)\,dx = 3.

Use the given information to evaluate ∫51[2f(x)−3g(x)]dx.\int_5^1 \left[2f(x) - 3g(x)\right] dx.

Hint

First apply linearity to split the integral over [1,5][1,5] into pieces involving ∫15f\int_1^5 f and ∫15g\int_1^5 g, then use the bounds-reversal property to switch from ∫51\int_5^1 to ∫15\int_1^5.

Step-by-step solution

Since reversing the limits of integration negates the integral, ∫51[2f(x)−3g(x)] dx=−∫15[2f(x)−3g(x)] dx\int_5^1 [2f(x)-3g(x)]\,dx = -\int_1^5 [2f(x)-3g(x)]\,dx. This lets us work with the given integrals over [1,5][1,5] instead.

By linearity, ∫15[2f(x)−3g(x)] dx=2∫15f(x) dx−3∫15g(x) dx=2(8)−3(3)=7\int_1^5 [2f(x)-3g(x)]\,dx = 2\int_1^5 f(x)\,dx - 3\int_1^5 g(x)\,dx = 2(8) - 3(3) = 7. Applying the negative sign from the bounds reversal gives ∫51[2f(x)−3g(x)] dx=−7\int_5^1[2f(x)-3g(x)]\,dx = -7.

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