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To evaluate a definite integral we must first integrate the function (we do not need to include c, the constant of integration) and then substitute in the given limits.

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(a)

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$\\int_\\var{a}^\\var{b}(1 + \\var{c}x)\\mathrm{dx} = \\left[x + \\var{c/2}x^2\\right]_\\var{a}^\\var{b}= [(\\var{b})+ \\var{c/2}(\\var{b})^2]-[(\\var{a}) + \\var{c/2}(\\var{a})^2]=\\simplify{{b}+ {c/2}{b}^2-{a} - {c/2}{a}^2}$

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(b)

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$\\int_\\var{d}^\\var{f} (x^2 + \\var{g}x-\\var{h})\\mathrm{dx}= \\left[\\frac{x^3}{3} + \\var{g/2}x^2-\\var{h}x\\right]_\\var{d}^\\var{f}=[\\frac{(\\var{f})^3}{3} + \\var{g/2}(\\var{f})^2-\\var{h}(\\var{f})]-[\\frac{(\\var{d})^3}{3} + \\var{g/2}(\\var{d})^2-\\var{h}(\\var{d})]=\\var{f^3/3 + g/2*f^2-h*f-(d^3/3 + g/2*d^2-h*d)}$

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(c)

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$\\int_\\var{j}^\\var{k}(x^3-\\var{l}x^2)\\mathrm{dx}=\\left[\\frac{x^4}{4} - \\frac{\\var{l}}{3}x^3\\right]_\\var{j}^\\var{k}=[\\frac{(\\var{k})^4}{4} - \\frac{\\var{l}}{3}(\\var{k})^3]-[\\frac{(\\var{j})^4}{4} - \\frac{\\var{l}}{3}(\\var{j})^3]=\\var{(k^4/4 - l/3*k^3)-(j^4/4 - l/3*j^3)}$

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$\\int_\\var{a}^\\var{b}(1 + \\var{c}x)\\mathrm{dx}$

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$\\int_\\var{d}^\\var{f} (x^2 + \\var{g}x-\\var{h})\\mathrm{dx}$

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$\\int_\\var{j}^\\var{k}(x^3-\\var{l}x^2)\\mathrm{dx}$

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Evaluate the following definite integrals, giving your answer as a fraction as necessary.

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Definite Integrals

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rebel

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rebelmaths

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Rebelmaths

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