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_{1}^{2} (x^{3} - 1) d x

\int_{- 2}^{1} (\frac{1}{3} t - 2)^{2} d t

\int_{0}^{1} e^{2} d x

\int_{0}^{1000} 0.03 x + 0.1 d x

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