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Using additivity to find a subinterval integral

Suppose ∫05f(x) dx=8\int_0^5 f(x)\,dx = 8 and ∫35f(x) dx=2\int_3^5 f(x)\,dx = 2.

Using the given information and the additivity property ∫acf(x) dx=∫abf(x) dx+∫bcf(x) dx\int_a^c f(x)\,dx = \int_a^b f(x)\,dx + \int_b^c f(x)\,dx, find ∫03f(x) dx\int_0^3 f(x)\,dx.

Hint

Split ∫05f(x) dx\int_0^5 f(x)\,dx into two adjacent pieces at x=3x=3, then solve for the unknown piece.

Step-by-step solution

By additivity over adjacent intervals, ∫05f(x) dx=∫03f(x) dx+∫35f(x) dx\int_0^5 f(x)\,dx = \int_0^3 f(x)\,dx + \int_3^5 f(x)\,dx, since 0<3<50 < 3 < 5.

Substituting the known values gives 8=∫03f(x) dx+28 = \int_0^3 f(x)\,dx + 2, so ∫03f(x) dx=8−2=6\int_0^3 f(x)\,dx = 8 - 2 = 6.

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