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Evaluating a combined integral using linearity and bounds-reversal

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

Using the given information, evaluate ∫51(4f(x)−2g(x))dx.\int_5^1 \left(4f(x) - 2g(x)\right)dx.

Hint

First use linearity to split the integral into separate pieces involving ff and gg, pulling out the constants. Then use the fact that reversing the limits of integration negates the integral's value.

Step-by-step solution

By linearity, ∫51(4f(x)−2g(x))dx=4∫51f(x) dx−2∫51g(x) dx\int_5^1 \left(4f(x)-2g(x)\right)dx = 4\int_5^1 f(x)\,dx - 2\int_5^1 g(x)\,dx. This splits the combined integral into two separate integrals with constants factored out front.

Reversing the bounds gives ∫51f(x) dx=−∫15f(x) dx=−8\int_5^1 f(x)\,dx = -\int_1^5 f(x)\,dx = -8 and ∫51g(x) dx=−∫15g(x) dx=−3\int_5^1 g(x)\,dx = -\int_1^5 g(x)\,dx = -3. Substituting these values: 4(−8)−2(−3)=−32+6=−264(-8) - 2(-3) = -32 + 6 = -26.

Sign change from reversing integration bounds

Hint
Step-by-step solution

Reversing the bounds of integration negates the integral's value, which is the same as multiplying by negative one. This is exactly the property used in part a, where integrating from five to one instead of one to five flipped the sign of each integral before combining them.

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