What is the integral from 0 to 4 of (x^2-2y^2+1) ?

The integral from 0 to 4 of (x^2-2y^2+1) is 4+64/3-8y^2

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integral from 0 to 4 of (x^2-2y^2+1)

\int_{0}^{4} (x^{2} - 2 y^{2} + 1) d x

4 + \frac{64}{3} - 8 y^{2}

Solution steps

\int_{0}^{4} x^{2} - 2 y^{2} + 1 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}^{4} 1 d x + \int_{0}^{4} x^{2} d x - \int_{0}^{4} 2 y^{2} d x

\int_{0}^{4} 1 d x = 4

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} 2 y^{2} d x = 8 y^{2}

= 4 + \frac{64}{3} - 8 y^{2}

\int_{2}^{5} 5^{5 x} d x

\int_{2}^{4} \frac{x}{2} d x

\int_{0}^{5} x (x^{2} - 1)^{3} d x

\int_{- 1}^{1} \frac{1}{x ^{2} + 2 x + 5} d x

\int_{0}^{\frac{1}{2}} \frac{1}{36 x ^{2} + 9} d x

What is the integral from 0 to 4 of (x^2-2y^2+1) ?
The integral from 0 to 4 of (x^2-2y^2+1) is 4+64/3-8y^2