∫ (from 0 to 1) of x-x^3 wrt x-এর সমাধান করুন

$$\int _ { 0 } ^ { 1 } x - x ^ { 3 } d x$$

মূল্যায়ন করুন

$$\frac{1}{4}=0.25$$

সমাধানের ধাপগুলো

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

$$\int x\mathrm{d}x+\int -x^{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

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

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}.

$$\frac{x^{2}}{2}-\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. -1 কে \frac{x^{4}}{4} বার গুণ করুন।

$$\frac{x^{2}}{2}-\frac{x^{4}}{4}$$

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.

$$\frac{1^{2}}{2}-\frac{1^{4}}{4}-\left(\frac{0^{2}}{2}-\frac{0^{4}}{4}\right)$$

সিমপ্লিফাই।

$$\frac{1}{4}$$