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

$$\int _ { 0 } ^ { 1 } ( x + 1 ) d x =$$

মূল্যায়ন করুন

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

সমাধানের ধাপগুলো

Evaluate the indefinite integral first.

$$\int x+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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}.

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

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

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

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.

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

সিমপ্লিফাই।

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