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

The integral of (x^4-1/2 x^3+1/4 x-2) is (x^5)/5-(x^4)/8+(x^2)/8-2x+C

Popular >

integral of (x^4-1/2 x^3+1/4 x-2)

\int (x^{4} - \frac{1}{2} x^{3} + \frac{1}{4} x - 2) d x

\frac{x ^{5}}{5} - \frac{x ^{4}}{8} + \frac{x ^{2}}{8} - 2 x + C

Solution steps

\int x^{4} - \frac{1}{2} x^{3} + \frac{1}{4} 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

= \int x^{4} d x - \int \frac{1}{2} x^{3} d x + \int \frac{1}{4} x d x - \int 2 d x

\int x^{4} d x = \frac{x ^{5}}{5}

\int \frac{1}{2} x^{3} d x = \frac{x ^{4}}{8}

\int \frac{1}{4} x d x = \frac{x ^{2}}{8}

\int 2 d x = 2 x

= \frac{x ^{5}}{5} - \frac{x ^{4}}{8} + \frac{x ^{2}}{8} - 2 x

Add a constant to the solution

= \frac{x ^{5}}{5} - \frac{x ^{4}}{8} + \frac{x ^{2}}{8} - 2 x + C

n = 1 \sum \infty (\frac{4 π}{5})^{n}

\int x^{2} + x - 3 d x

\frac{d}{d x} (sin (4 x) + sin^{2} (4 x))

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

\frac{d ^{2} x}{d t} + 2 \frac{d x}{d t} + x = 1

What is the integral of (x^4-1/2 x^3+1/4 x-2) ?
The integral of (x^4-1/2 x^3+1/4 x-2) is (x^5)/5-(x^4)/8+(x^2)/8-2x+C