∫ (from 0 to 1) of 2^x+x^2 wrt x-এর সমাধান করুন

$$\int _ { 0 } ^ { 1 } 2 ^ { x } + x ^ { 2 } d x$$

মূল্যায়ন করুন

$$\frac{1}{\ln(2)}+\frac{1}{3}\approx 1.776028374$$

সমাধানের ধাপগুলো

Evaluate the indefinite integral first.

$$\int 2^{x}+x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2^{x}\mathrm{d}x+\int x^{2}\mathrm{d}x$$

Use \int x^{k}\mathrm{d}k=\frac{x^{k}}{\ln(x)} from the table of common integrals to obtain the result.

$$\frac{2^{x}}{\ln(2)}+\int x^{2}\mathrm{d}x$$

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}.

$$\frac{2^{x}}{\ln(2)}+\frac{x^{3}}{3}$$

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.

$$2^{1}\ln(2)^{-1}+\frac{1^{3}}{3}-\left(2^{0}\ln(2)^{-1}+\frac{0^{3}}{3}\right)$$

সিমপ্লিফাই।

$$\frac{1}{3}+\frac{1}{\ln(2)}$$