# Evaluate the integral. \int_{0}^{1}(4x-9x^{2})dx

Evaluate the integral.
$$\displaystyle{\int_{{{0}}}^{{{1}}}}{\left({4}{x}-{9}{x}^{{{2}}}\right)}{\left.{d}{x}\right.}$$

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Ralph Lester
Step 1
Solve the integral $$\displaystyle\int{\left({4}{x}-{9}{x}^{{{2}}}\right)}{\left.{d}{x}\right.}$$ by using the general power rule of the integration that is $$\displaystyle\int{x}^{{{n}}}={\frac{{{x}^{{{n}+{1}}}}}{{{n}+{1}}}}{\left.{d}{x}\right.}$$, where n is not equal to 1.
$$\displaystyle\int{\left({4}{x}-{9}{x}^{{{2}}}\right)}{\left.{d}{x}\right.}={\frac{{{4}{x}^{{{1}+{1}}}}}{{{1}+{1}}}}-{\frac{{{9}{x}^{{{2}+{1}}}}}{{{2}+{1}}}}$$
$$\displaystyle={\frac{{{4}{x}^{{{2}}}}}{{{2}}}}-{\frac{{{9}{x}^{{{3}}}}}{{{3}}}}$$
$$\displaystyle={2}{x}^{{{2}}}-{3}{x}^{{{3}}}$$
Step 2
Now, use the fundamental theorem of calculus and apply the upper and lower limit 1 and 0 in the integral solution of the integral $$\displaystyle{\int_{{{0}}}^{{{1}}}}{\left({4}{x}-{9}{x}^{{{2}}}\right)}{\left.{d}{x}\right.}$$ to calculate the final value.
$$\displaystyle{\int_{{{0}}}^{{{1}}}}{\left({4}{x}-{9}{x}^{{{2}}}\right)}{\left.{d}{x}\right.}={2}{x}^{{{2}}}-{3}{x}^{{{3}}}{{\mid}_{{{0}}}^{{{1}}}}$$
$$\displaystyle={2}{\left({1}\right)}^{{{2}}}-{3}{\left({1}\right)}^{{{3}}}-{\left[{2}{\left({0}\right)}^{{{2}}}-{3}{\left({0}\right)}^{{{3}}}\right]}$$
=2-3-0
=-1
###### Have a similar question?
George Morin
Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:
$$\displaystyle{\int_{{{a}}}^{{{b}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}{{\mid}_{{{a}}}^{{{b}}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$
Step 2: In this case, $$\displaystyle{f{{\left({x}\right)}}}={\left({4}{x}-{9}{x}^{{{2}}}\right)}$$. Find its integral.
$$\displaystyle{2}{x}^{{{2}}}-{3}{x}^{{{3}}}{{\mid}_{{{0}}}^{{{1}}}}$$
Step 3: Since $$\displaystyle{F}{\left({x}\right)}{{\mid}_{{{a}}}^{{{b}}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$, expand the above into F(1)−F(0):
$$\displaystyle{\left({2}\times{1}^{{{2}}}-{3}\times{1}^{{{3}}}\right)}-{\left({2}\times{0}^{{{2}}}-{3}\times{0}^{{{3}}}\right)}$$
Step 4: Simplify.
-1