Solve ∫ 2pi/60x+17,5xpi wrt x

$$\int \frac { 2 \pi } { 60 } x + 17,5 x \pi d x$$

Evaluate

$$\frac{263\pi x^{2}}{30}+С$$

Solution Steps

Divide 2\pi by 60 to get \frac{1}{30}\pi .

$$\int \frac{1}{30}\pi x+17,5x\pi \mathrm{d}x$$

Combine \frac{1}{30}\pi x and 17,5x\pi to get \frac{263}{15}\pi x.

$$\int \frac{263}{15}\pi x\mathrm{d}x$$

Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.

$$\frac{263\pi }{15}\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}.

$$\frac{263\pi }{15}\times \frac{x^{2}}{2}$$

Simplify.

$$\frac{263\pi x^{2}}{30}$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{263\pi x^{2}}{30}+С$$

Differentiate w.r.t. x

$$\frac{263\pi x}{15}$$