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

$$\int _ { 0 } ^ { 4 } x ^ { 2 } + 4 x d x$$

মূল্যায়ন করুন

$$\frac{160}{3}\approx 53.333333333$$

সমাধানের ধাপগুলো

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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}. 4 কে \frac{x^{2}}{2} বার গুণ করুন।

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

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.

$$\frac{4^{3}}{3}+2\times 4^{2}-\left(\frac{0^{3}}{3}+2\times 0^{2}\right)$$

সিমপ্লিফাই।

$$\frac{160}{3}$$