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

$$\int _ { 1 } ^ { 2 } 2 x ^ { 2 } + 3 x - 3 d x$$

মূল্যায়ন করুন

$$\frac{37}{6}\approx 6.166666667$$

সমাধানের ধাপগুলো

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}. 2 কে \frac{x^{3}}{3} বার গুণ করুন।

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

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}. 3 কে \frac{x^{2}}{2} বার গুণ করুন।

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

Find the integral of -3 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

সিমপ্লিফাই।

$$\frac{37}{6}$$