∫ (from 2 to 4) of 3x^5-2x^3+x-এর সমাধান করুন

$$\int _ { 2 } ^ { 4 } 3 x ^ { 5 } - 2 x ^ { 3 } + x$$

মূল্যায়ন করুন

$$1902$$

সমাধানের ধাপগুলো

Evaluate the indefinite integral first.

$$\int 3x^{5}-2x^{3}+x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x^{5}\mathrm{d}x+\int -2x^{3}\mathrm{d}x+\int x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x^{5}\mathrm{d}x-2\int x^{3}\mathrm{d}x+\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}. 3 কে \frac{x^{6}}{6} বার গুণ করুন।

$$\frac{x^{6}}{2}-2\int x^{3}\mathrm{d}x+\int x\mathrm{d}x$$

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}. -2 কে \frac{x^{4}}{4} বার গুণ করুন।

$$\frac{x^{6}}{2}-\frac{x^{4}}{2}+\int x\mathrm{d}x$$

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}.

$$\frac{x^{6}-x^{4}+x^{2}}{2}$$

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.

$$\frac{4^{6}}{2}-\frac{4^{4}}{2}+\frac{4^{2}}{2}-\left(\frac{2^{6}}{2}-\frac{2^{4}}{2}+\frac{2^{2}}{2}\right)$$

সিমপ্লিফাই।

$$1902$$