f ( x ) = ∫ (from 0 to x) of (t^3+2t^2+1)dt-এর সমাধান করুন

$$f ( x ) = \int _ { 0 } ^ { x } ( t ^ { 3 } + 2 t ^ { 2 } + 1 ) d t$$

মূল্যায়ন করুন

$$\frac{x^{4}}{4}+\frac{2x^{3}}{3}+x$$

সমাধানের ধাপগুলো

Evaluate the indefinite integral first.

$$\int t^{3}+2t^{2}+1\mathrm{d}t$$

Integrate the sum term by term.

$$\int t^{3}\mathrm{d}t+\int 2t^{2}\mathrm{d}t+\int 1\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\int t^{3}\mathrm{d}t+2\int t^{2}\mathrm{d}t+\int 1\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{3}\mathrm{d}t with \frac{t^{4}}{4}.

$$\frac{t^{4}}{4}+2\int t^{2}\mathrm{d}t+\int 1\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{2}\mathrm{d}t with \frac{t^{3}}{3}. 2 কে \frac{t^{3}}{3} বার গুণ করুন।

$$\frac{t^{4}}{4}+\frac{2t^{3}}{3}+\int 1\mathrm{d}t$$

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

$$\frac{t^{4}}{4}+\frac{2t^{3}}{3}+t$$

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.

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

সিমপ্লিফাই।

$$\frac{x^{4}}{4}+\frac{2x^{3}}{3}+x$$

w.r.t. x পার্থক্য করুন

$$x^{3}+2x^{2}+1$$