∫ (from 1 to 6) of x^3+5x wrt x-এর সমাধান করুন

$$\int _ { 1 } ^ { 6 } x ^ { 3 } + 5 x d x$$

মূল্যায়ন করুন

$$\frac{1645}{4}=411.25$$

সমাধানের ধাপগুলো

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

$$\frac{x^{4}}{4}+5\int x\mathrm{d}x$$

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}. 5 কে \frac{x^{2}}{2} বার গুণ করুন।

$$\frac{x^{4}}{4}+\frac{5x^{2}}{2}$$

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.

$$\frac{6^{4}}{4}+\frac{5}{2}\times 6^{2}-\left(\frac{1^{4}}{4}+\frac{5}{2}\times 1^{2}\right)$$

সিমপ্লিফাই।

$$\frac{1645}{4}$$