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\int x^{2}-6x+5\mathrm{d}x
Evaluate the indefinite integral first.
\int x^{2}\mathrm{d}x+\int -6x\mathrm{d}x+\int 5\mathrm{d}x
Integrate the sum term by term.
\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x+\int 5\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{3}}{3}-6\int x\mathrm{d}x+\int 5\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}-3x^{2}+\int 5\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}. -6 কে \frac{x^{2}}{2} বার গুণ করুন।
\frac{x^{3}}{3}-3x^{2}+5x
Find the integral of 5 using the table of common integrals rule \int a\mathrm{d}x=ax.
\frac{1^{3}}{3}-3\times 1^{2}+5\times 1-\left(\frac{\left(-5\right)^{3}}{3}-3\left(-5\right)^{2}+5\left(-5\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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