What is the integral of (-0.98wlx-0.98wl^2) ?

The integral of (-0.98wlx-0.98wl^2) is -0.98l^2wx-0.49lwx^2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (-0.98wlx-0.98wl^2)

\int (- 0.98 wl x - 0.98 w l^{2}) d x

- 0.98 l^{2} w x - 0.49 lw x^{2} + C

Solution steps

\int - 0.98 wl x - 0.98 w l^{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 0.98 l^{2} w d x - \int 0.98 lw x d x

\int 0.98 l^{2} w d x = 0.98 l^{2} w x

\int 0.98 lw x d x = 0.49 lw x^{2}

= - 0.98 l^{2} w x - 0.49 lw x^{2}

Add a constant to the solution

= - 0.98 l^{2} w x - 0.49 lw x^{2} + C

3 sin (t) cos (t)

n = 1 \sum \infty (0.2)^{n}

\int_{\frac{π}{2}}^{π} cos^{2} (x) d x

2 y^{^{''}} + 18 y = 0

\int \frac{x}{( 3 - x ^{2} ) ^{6}} d x

What is the integral of (-0.98wlx-0.98wl^2) ?
The integral of (-0.98wlx-0.98wl^2) is -0.98l^2wx-0.49lwx^2+C