What is the integral from 0 to L of (3ax^2+2bx+c)^2 ?

The integral from 0 to L of (3ax^2+2bx+c)^2 is (9a^2L^5)/5+3abL^4+2acL^3+(4b^2L^3)/3+2bcL^2+c^2L

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integral from 0 to L of (3ax^2+2bx+c)^2

\int_{0}^{L} (3 a x^{2} + 2 b x + c)^{2} d x

\frac{9 a ^{2} L ^{5}}{5} + 3 ab L^{4} + 2 a c L^{3} + \frac{4 b ^{2} L ^{3}}{3} + 2 b c L^{2} + c^{2} L

Solution steps

\int_{0}^{L} (3 a x^{2} + 2 b x + c)^{2} d x

Expand

(3 a x^{2} + 2 b x + c)^{2} : 9 a^{2} x^{4} + 12 ab x^{3} + 6 a c x^{2} + 4 b^{2} x^{2} + 4 b c x + c^{2}

= \int_{0}^{L} 9 a^{2} x^{4} + 12 ab x^{3} + 6 a c x^{2} + 4 b^{2} x^{2} + 4 b c x + c^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{L} 9 a^{2} x^{4} d x + \int_{0}^{L} 12 ab x^{3} d x + \int_{0}^{L} 6 a c x^{2} d x + \int_{0}^{L} 4 b^{2} x^{2} d x + \int_{0}^{L} 4 b c x d x + \int_{0}^{L} c^{2} d x

\int_{0}^{L} 9 a^{2} x^{4} d x = \frac{9 a ^{2} L ^{5}}{5}

\int_{0}^{L} 12 ab x^{3} d x = 3 ab L^{4}

\int_{0}^{L} 6 a c x^{2} d x = 2 a c L^{3}

\int_{0}^{L} 4 b^{2} x^{2} d x = \frac{4 b ^{2} L ^{3}}{3}

\int_{0}^{L} 4 b c x d x = 2 b c L^{2}

\int_{0}^{L} c^{2} d x = c^{2} L

= \frac{9 a ^{2} L ^{5}}{5} + 3 ab L^{4} + 2 a c L^{3} + \frac{4 b ^{2} L ^{3}}{3} + 2 b c L^{2} + c^{2} L

What is the integral from 0 to L of (3ax^2+2bx+c)^2 ?
The integral from 0 to L of (3ax^2+2bx+c)^2 is (9a^2L^5)/5+3abL^4+2acL^3+(4b^2L^3)/3+2bcL^2+c^2L