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Popular >

integral from 0 to 4 of 16-x^3

Solution

\int_{0}^{4} 16 - x^{3} d x

Solution

0

Solution steps

\int_{0}^{4} 16 - x^{3} 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} 16 d x - \int_{0}^{4} x^{3} d x

\int_{0}^{4} 16 d x = 64

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

= 64 - 64

Simplify

= 0

Popular Examples

(dy)/(dx)-1/x*y-3*x^2=0

\frac{d y}{d x} - \frac{1}{x} \cdot y - 3 \cdot x^{2} = 0

(dy)/(dx)=((e^x))/(ysqrt(1+y^2))

\frac{d y}{d x} = \frac{( e ^{x} )}{y 1 + y ^{2}}

tangent of f(x)=(7x)/(x+4),\at x=3 tangent of

f (x) = \frac{7 x}{x + 4}, at x = 3

integral of sin^3(3x)cos^2(3x)

\int sin^{3} (3 x) cos^{2} (3 x) d x

tangent of f(x)=(6x)/(x+3),(-1,-3)tangent of

f (x) = \frac{6 x}{x + 3}, (- 1, - 3)

Frequently Asked Questions (FAQ)

What is the integral from 0 to 4 of 16-x^3 ?
The integral from 0 to 4 of 16-x^3 is 0