∫ (x-3)^3 wrt x-এর সমাধান করুন

$$\int ( x - 3 ) ^ { 3 } d x$$

মূল্যায়ন করুন

$$\frac{\left(x-3\right)^{4}}{4}+С$$

সমাধানের ধাপগুলো

\left(x-3\right)^{3} প্রসারিত করতে বাইনোমিয়াল উপপাদ্য \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} ব্যবহার করুন৷

$$\int x^{3}-9x^{2}+27x-27\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int -9x^{2}\mathrm{d}x+\int 27x\mathrm{d}x+\int -27\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x-9\int x^{2}\mathrm{d}x+27\int x\mathrm{d}x+\int -27\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}-9\int x^{2}\mathrm{d}x+27\int x\mathrm{d}x+\int -27\mathrm{d}x$$

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}. -9 কে \frac{x^{3}}{3} বার গুণ করুন।

$$\frac{x^{4}}{4}-3x^{3}+27\int x\mathrm{d}x+\int -27\mathrm{d}x$$

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

$$\frac{x^{4}}{4}-3x^{3}+\frac{27x^{2}}{2}+\int -27\mathrm{d}x$$

Find the integral of -27 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{4}-3x^{3}+\frac{27x^{2}}{2}-27x$$

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.

$$\frac{x^{4}}{4}-3x^{3}+\frac{27x^{2}}{2}-27x+С$$

w.r.t. x পার্থক্য করুন

$$\left(x-3\right)^{3}$$