∫ (from -2 to 1) of 3x^2+5x-1 wrt x-এর সমাধান করুন

$$\int_{ -2 }^{ 1 } 3 { x }^{ 2 } +5x-1 d x$$

মূল্যায়ন করুন

$$-\frac{3}{2}=-1.5$$

সমাধানের ধাপগুলো

Evaluate the indefinite integral first.

$$\int 3x^{2}+5x-1\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x^{2}\mathrm{d}x+\int 5x\mathrm{d}x+\int -1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x^{2}\mathrm{d}x+5\int x\mathrm{d}x+\int -1\mathrm{d}x$$

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}. 3 কে \frac{x^{3}}{3} বার গুণ করুন।

$$x^{3}+5\int x\mathrm{d}x+\int -1\mathrm{d}x$$

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।

$$x^{3}+\frac{5x^{2}}{2}+\int -1\mathrm{d}x$$

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$x^{3}+\frac{5x^{2}}{2}-x$$

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.

$$1^{3}+\frac{5}{2}\times 1^{2}-1-\left(\left(-2\right)^{3}+\frac{5}{2}\left(-2\right)^{2}-\left(-2\right)\right)$$

সিমপ্লিফাই।

$$-\frac{3}{2}$$