For a continuous function on a closed interval, evaluate an antiderivative at the upper and lower limits. A definite integral is a number; it has no arbitrary constant.
\(\int_a^b \mathrm{f}(x)\,\mathrm{d}x=[\mathrm{F}(x)]_a^b=\mathrm{F}(b)-\mathrm{F}(a)\).
\(\int_1^3(2x+1)\,\mathrm{d}x=[x^2+x]_1^3=12-2=10\).
| Property | Identity |
|---|---|
| Reversed limits | \(\int_b^a \mathrm{f}(x)\,\mathrm{d}x=-\int_a^b \mathrm{f}(x)\,\mathrm{d}x\) |
| Split an interval | \(\int_a^b \mathrm{f}(x)\,\mathrm{d}x=\int_a^c \mathrm{f}(x)\,\mathrm{d}x+\int_c^b \mathrm{f}(x)\,\mathrm{d}x\), \(a<c<b\) |
| Constant multiple | \(\int_a^b k\mathrm{f}(x)\,\mathrm{d}x=k\int_a^b \mathrm{f}(x)\,\mathrm{d}x\) |
With \(u=x^2+1\), \(\mathrm{d}u=2x\,\mathrm{d}x\), the limits \(x=0,1\) become \(u=1,2\):
\(\int_0^1 2x(x^2+1)^3\,\mathrm{d}x=\int_1^2u^3\,\mathrm{d}u=\left[\dfrac{u^4}{4}\right]_1^2=\dfrac{15}{4}\).
Use limits in the same variable as the integrand. Alternatively substitute back to the original variable before using the original limits.
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