∫ (from 0 to 100) of (x^2+2+3x^4+2e^9) wrt x-এর সমাধান করুন

$$\int _ { 0 } ^ { 100 } ( x ^ { 2 } + 2 + 3 x ^ { 4 } + 2 e ^ { 9 } ) d x$$

মূল্যায়ন করুন

$$200e^{9}+\frac{18001000600}{3}\approx 6001954150.118847847$$

সমাধানের ধাপগুলো

Evaluate the indefinite integral first.

$$\int x^{2}+2+3x^{4}+2e^{9}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int 2\mathrm{d}x+\int 3x^{4}\mathrm{d}x+\int 2e^{9}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x+\int 2\mathrm{d}x+3\int x^{4}\mathrm{d}x+2\int e^{9}\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}+\int 2\mathrm{d}x+3\int x^{4}\mathrm{d}x+2\int e^{9}\mathrm{d}x$$

Find the integral of 2 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}+2x+3\int x^{4}\mathrm{d}x+2\int e^{9}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. 3 কে \frac{x^{5}}{5} বার গুণ করুন।

$$\frac{x^{3}}{3}+2x+\frac{3x^{5}}{5}+2\int e^{9}\mathrm{d}x$$

Find the integral of e^{9} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}+2x+\frac{3x^{5}}{5}+2e^{9}x$$

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.

$$\frac{100^{3}}{3}+2\times 100+\frac{3}{5}\times 100^{5}+2e^{9}\times 100-\left(\frac{0^{3}}{3}+2\times 0+\frac{3}{5}\times 0^{5}+2e^{9}\times 0\right)$$

সিমপ্লিফাই।

$$\frac{18001000600}{3}+200e^{9}$$