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\int 3x^{2}+x-1\mathrm{d}x
Evaluate the indefinite integral first.
\int 3x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int -1\mathrm{d}x
Integrate the sum term by term.
3\int x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int -1\mathrm{d}x
Factor out the constant in each of the terms.
x^{3}+\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}+\frac{x^{2}}{2}+\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}.
x^{3}+\frac{x^{2}}{2}-x
Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}x=ax.
0^{3}+\frac{0^{2}}{2}-0-\left(\left(-2\right)^{3}+\frac{\left(-2\right)^{2}}{2}-\left(-2\right)\right)
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.
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