What is the integral of (1-x^2)/(1600) ?

The integral of (1-x^2)/(1600) is 1/1600 (x-(x^3)/3)+C

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integral of (1-x^2)/(1600)

\int \frac{1 - x ^{2}}{1600} d x

\frac{1}{1600} (x - \frac{x ^{3}}{3}) + C

Solution steps

\int \frac{1 - x ^{2}}{1600} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{1600} \cdot \int 1 - x^{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

= \frac{1}{1600} (\int 1 d x - \int x^{2} d x)

\int 1 d x = x

\int x^{2} d x = \frac{x ^{3}}{3}

= \frac{1}{1600} (x - \frac{x ^{3}}{3})

Add a constant to the solution

= \frac{1}{1600} (x - \frac{x ^{3}}{3}) + C

x ln (x^{2})

y^{^{''}} + 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = - x^{2} + 7 x, y = x^{2} - 3 x

n = 5 \sum \infty \frac{6 ^{n}}{7 ^{n}}

x \to 0 + lim ((5 x)^{4 x})

What is the integral of (1-x^2)/(1600) ?
The integral of (1-x^2)/(1600) is 1/1600 (x-(x^3)/3)+C