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Splitting an integral via linearity

Given the known values ∫04x2 dx=643\int_0^4 x^2\,dx = \dfrac{64}{3}, ∫04x dx=8\int_0^4 x\,dx = 8, and ∫041 dx=4\int_0^4 1\,dx = 4, consider the function f(x)=3x2−2x+1f(x) = 3x^2 - 2x + 1 on [0,4][0,4].

Use linearity of the definite integral to rewrite ∫04(3x2−2x+1) dx\int_0^4 (3x^2 - 2x + 1)\,dx as a combination of the three known integrals ∫04x2 dx\int_0^4 x^2\,dx, ∫04x dx\int_0^4 x\,dx, and ∫041 dx\int_0^4 1\,dx given above, and evaluate the result to find ∫04(3x2−2x+1) dx\int_0^4 (3x^2 - 2x + 1)\,dx.

Hint

Break the integral into separate terms using the sum/difference and constant-multiple rules for integrals, then plug in the three given values.

Step-by-step solution

By linearity, ∫04(3x2−2x+1) dx=3∫04x2 dx−2∫04x dx+∫041 dx\int_0^4 (3x^2 - 2x + 1)\,dx = 3\int_0^4 x^2\,dx - 2\int_0^4 x\,dx + \int_0^4 1\,dx. This splits the single integral into three simpler pieces using the constant-multiple and sum/difference properties.

Substituting the given values: 3(643)−2(8)+4=64−16+4=523\left(\dfrac{64}{3}\right) - 2(8) + 4 = 64 - 16 + 4 = 52. So ∫04(3x2−2x+1) dx=52\int_0^4 (3x^2 - 2x + 1)\,dx = 52.

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