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

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

মূল্যায়ন করুন

$$48$$

সমাধানের ধাপগুলো

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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।

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

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.

$$\frac{3^{4}}{4}+3^{3}-\left(\frac{\left(-1\right)^{4}}{4}+\left(-1\right)^{3}\right)$$

সিমপ্লিফাই।

$$48$$