Solve ∫ (from 0 to 2) of 5x^2+u^2

$$\int _ { 0 } ^ { 2 } 5 x ^ { 2 } + u ^ { 2 }$$

Evaluate

$$2u^{2}+\frac{40}{3}$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 5x^{2}+u^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 5x^{2}\mathrm{d}x+\int u^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$5\int x^{2}\mathrm{d}x+\int u^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 5 times \frac{x^{3}}{3}.

$$\frac{5x^{3}}{3}+\int u^{2}\mathrm{d}x$$

Find the integral of u^{2} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{5x^{3}}{3}+u^{2}x$$

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.

$$\frac{5}{3}\times 2^{3}+u^{2}\times 2-\left(\frac{5}{3}\times 0^{3}+u^{2}\times 0\right)$$

Simplify.

$$\frac{40}{3}+2u^{2}$$

Differentiate w.r.t. u

$$4u$$