Evaluate
\frac{4}{3}\approx 1.333333333
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\int 1-x^{2}\mathrm{d}x
Evaluate the indefinite integral first.
\int 1\mathrm{d}x+\int -x^{2}\mathrm{d}x
Integrate the sum term by term.
\int 1\mathrm{d}x-\int x^{2}\mathrm{d}x
Factor out the constant in each of the terms.
x-\int x^{2}\mathrm{d}x
Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.
x-\frac{x^{3}}{3}
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}. Multiply -1 times \frac{x^{3}}{3}.
1-\frac{1^{3}}{3}-\left(-1-\frac{\left(-1\right)^{3}}{3}\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.
\frac{4}{3}
Simplify.
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\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
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Limits
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