∫ 1.2x^2+16.2x+62.3 wrt x-এর সমাধান করুন

$$\int 1.2 x ^ { 2 } + 16.2 x + 62.3 d x$$

মূল্যায়ন করুন

$$\frac{2x^{3}}{5}+\frac{81x^{2}}{10}+\frac{623x}{10}+С$$

সমাধানের ধাপগুলো

Integrate the sum term by term.

$$\int \frac{6x^{2}}{5}\mathrm{d}x+\int \frac{81x}{5}\mathrm{d}x+\int 62.3\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{6\int x^{2}\mathrm{d}x}{5}+\frac{81\int x\mathrm{d}x}{5}+\int 62.3\mathrm{d}x$$

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

$$\frac{2x^{3}}{5}+\frac{81\int x\mathrm{d}x}{5}+\int 62.3\mathrm{d}x$$

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

$$\frac{2x^{3}}{5}+\frac{81x^{2}}{10}+\int 62.3\mathrm{d}x$$

Find the integral of 62.3 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{2x^{3}}{5}+\frac{81x^{2}}{10}+\frac{623x}{10}$$

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.

$$\frac{2x^{3}}{5}+\frac{81x^{2}}{10}+\frac{623x}{10}+С$$

w.r.t. x পার্থক্য করুন

$$\frac{6x^{2}}{5}+\frac{81x}{5}+62.3$$