Լուծեք ∫ _{frac{sqrt{2}}{2}}^1(x^4-1/2x^2) wrt x

$$\int _ { \frac { \sqrt { 2 } } { 2 } } ^ { 1 } ( x ^ { 4 } - \frac { 1 } { 2 } x ^ { 2 } ) d x$$

Գնահատել

$$\frac{\sqrt{2}}{60}+\frac{1}{30}\approx 0.056903559$$

Լուծման քայլերը

Evaluate the indefinite integral first.

$$\int x^{4}-\frac{x^{2}}{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{4}\mathrm{d}x+\int -\frac{x^{2}}{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{4}\mathrm{d}x-\frac{\int x^{2}\mathrm{d}x}{2}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}.

$$\frac{x^{5}}{5}-\frac{\int x^{2}\mathrm{d}x}{2}$$

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}. Բազմապատկեք -\frac{1}{2} անգամ \frac{x^{3}}{3}:

$$\frac{x^{5}}{5}-\frac{x^{3}}{6}$$

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.

$$\frac{1^{5}}{5}-\frac{1^{3}}{6}-\left(\frac{1}{5}\times \left(\frac{1}{2}\times 2^{\frac{1}{2}}\right)^{5}-\frac{1}{6}\times \left(\frac{1}{2}\times 2^{\frac{1}{2}}\right)^{3}\right)$$

Պարզեցնել:

$$\frac{1}{30}+\frac{\sqrt{2}}{60}$$