Solve ∫ (from 67 to r ^ 2) of +3x

$$\int _ { 67 } ^ { r ^ { 2 } } + 3 x$$

Evaluate

$$\frac{3\left(r^{4}-4489\right)}{2}$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 3x\mathrm{d}x$$

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

$$3\int x\mathrm{d}x$$

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}.

$$\frac{3x^{2}}{2}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$\frac{3}{2}\left(r^{2}\right)^{2}-\frac{3}{2}\times 67^{2}$$

Simplify.

$$\frac{3r^{4}-13467}{2}$$

Differentiate w.r.t. r

$$6r^{3}$$