What is the integral from 1 to 3 of kx^2-2 ?

The integral from 1 to 3 of kx^2-2 is (26k)/3-4

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integral from 1 to 3 of kx^2-2

\int_{1}^{3} k x^{2} - 2 d x

\frac{26 k}{3} - 4

Solution steps

\int_{1}^{3} k x^{2} - 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_{1}^{3} k x^{2} d x - \int_{1}^{3} 2 d x

\int_{1}^{3} k x^{2} d x = \frac{26 k}{3}

\int_{1}^{3} 2 d x = 4

= \frac{26 k}{3} - 4

\int_{0}^{2} - 2 x + 2 d x

\int_{0}^{1} \frac{1}{1 + 2 x} d x

\int_{0}^{6 - x - z} 1 d y

\int_{0}^{π} t sin (2 t) d t

\int_{2}^{6} (3 + 2 x) d x

What is the integral from 1 to 3 of kx^2-2 ?
The integral from 1 to 3 of kx^2-2 is (26k)/3-4