What is the integral of 1-x^2+1/2 x^4 ?

The integral of 1-x^2+1/2 x^4 is x-(x^3)/3+(x^5)/(10)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1-x^2+1/2 x^4

\int 1 - x^{2} + \frac{1}{2} x^{4} d x

x - \frac{x ^{3}}{3} + \frac{x ^{5}}{10} + C

Solution steps

\int 1 - x^{2} + \frac{1}{2} x^{4} 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 d x - \int x^{2} d x + \int \frac{1}{2} x^{4} d x

\int 1 d x = x

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

\int \frac{1}{2} x^{4} d x = \frac{x ^{5}}{10}

= x - \frac{x ^{3}}{3} + \frac{x ^{5}}{10}

Add a constant to the solution

= x - \frac{x ^{3}}{3} + \frac{x ^{5}}{10} + C

x^{3} - 2 x^{2} + 3 x - 5, 2

x \to 4 lim (2 x^{2} - 11 x)

\frac{d}{d x} (\frac{42}{x ^{4}})

x \to 0 lim (x^{2} + 3)

y^{^{'}} x^{2} + 2 x y = 5 y^{3}

What is the integral of 1-x^2+1/2 x^4 ?
The integral of 1-x^2+1/2 x^4 is x-(x^3)/3+(x^5)/(10)+C