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Find the area under the curve f(x)=x^{3}+4x^{2}-7x+8 on the interval [0,1].

If you look at this function on a graphing calculator, you will see that it is always positive on the interval [0,1]. Thus, we need only one integral to find the area.

A=\int _{{0}}^{{1}}(x^{3}+4x^{2}-7x+8)\,dx=\left[{\frac  {1}{4}}x^{4}+{\frac  {4}{3}}x^{3}-{\frac  {7}{2}}x^{2}+8x\right]_{{0}}^{{1}}={\frac  {1}{4}}+{\frac  {4}{3}}-{\frac  {7}{2}}+8={\frac  {73}{12}}\,

Thus, the area between the graph and the x-axis on the interval [0,1] is {\frac  {73}{12}} square units.

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