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

$$\int_{ 1 }^{ 4 } { x }^{ 2 } + \frac{ 1 }{ { x }^{ 3 } } d x$$

মূল্যায়ন করুন

$$\frac{687}{32}=21.46875$$

সমাধানের ধাপগুলো

Evaluate the indefinite integral first.

$$\int x^{2}+\frac{1}{x^{3}}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int \frac{1}{x^{3}}\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}+\int \frac{1}{x^{3}}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int \frac{1}{x^{3}}\mathrm{d}x with -\frac{1}{2x^{2}}.

$$\frac{x^{3}}{3}-\frac{1}{2x^{2}}$$

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.

$$\frac{4^{3}}{3}-\frac{4^{-2}}{2}-\left(\frac{1^{3}}{3}-\frac{1^{-2}}{2}\right)$$

সিমপ্লিফাই।

$$\frac{687}{32}$$