∫ (1/4x^3-1/3x^2+1/2x) wrt x-এর সমাধান করুন

$$\int ( \frac { 1 } { 4 } x ^ { 3 } - \frac { 1 } { 3 } x ^ { 2 } + \frac { 1 } { 2 } x ) d x$$

মূল্যায়ন করুন

$$\frac{x^{4}}{16}-\frac{x^{3}}{9}+\frac{x^{2}}{4}+С$$

সমাধানের ধাপগুলো

Integrate the sum term by term.

$$\int \frac{x^{3}}{4}\mathrm{d}x+\int -\frac{x^{2}}{3}\mathrm{d}x+\int \frac{x}{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{\int x^{3}\mathrm{d}x}{4}-\frac{\int x^{2}\mathrm{d}x}{3}+\frac{\int x\mathrm{d}x}{2}$$

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}. \frac{1}{4} কে \frac{x^{4}}{4} বার গুণ করুন।

$$\frac{x^{4}}{16}-\frac{\int x^{2}\mathrm{d}x}{3}+\frac{\int x\mathrm{d}x}{2}$$

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}. -\frac{1}{3} কে \frac{x^{3}}{3} বার গুণ করুন।

$$\frac{x^{4}}{16}-\frac{x^{3}}{9}+\frac{\int x\mathrm{d}x}{2}$$

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{1}{2} কে \frac{x^{2}}{2} বার গুণ করুন।

$$\frac{x^{4}}{16}-\frac{x^{3}}{9}+\frac{x^{2}}{4}$$

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.

$$\frac{x^{4}}{16}-\frac{x^{3}}{9}+\frac{x^{2}}{4}+С$$

w.r.t. x পার্থক্য করুন

$$\frac{x^{3}}{4}-\frac{x^{2}}{3}+\frac{x}{2}$$