How do you evaluate the integral $\int x^{3} + 4 x^{2} + 5 d x$ $intx^3+4x^2+5 dx$ ?

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How do you evaluate the integral $\int x^{3} + 4 x^{2} + 5 d x$ $intx^3+4x^2+5 dx$ ?

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Answer 1 · 2014-08-24T19:09:41

Because this equation only consists of terms added together, you can integrate them separately and add the results, giving us:

$\int x^{3} + 4 x^{2} + 5 d x = \int x^{3} d x + \int 4 x^{2} d x + \int 5 d x$ $int x^3 + 4x^2 + 5dx = intx^3dx + int4x^2dx + int5dx$

Each of these terms can be integrated using the Power Rule for integration, which is:

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $int x^ndx = x^(n+1)/(n+1) + C$

Plugging our 3 terms into this formula, we have:

$\int x^{3} d x = \frac{x^{3 + 1}}{3 + 1} = \frac{x^{4}}{4}$ $int x^3dx = x^(3+1)/(3+1) = x^4/4$

$\int 4 x^{2} d x = \frac{4 x^{2 + 1}}{2 + 1} = \frac{4 x^{3}}{3}$ $int 4x^2dx = (4x^(2+1))/(2+1) = (4x^3)/3$

$\int 5 d x = \int 5 x^{0} d x = \frac{5 x^{0 + 1}}{0 + 1} = \frac{5 x^{1}}{1} = 5 x$ $int 5dx = int 5x^0dx = (5x^(0+1))/(0+1) = (5x^1)/1 = 5x$

Now we arrive at our final answer by adding these together, remembering to add our constant ( $C$ $C$ ) on the end:

$\int x^{3} + 4 x^{2} + 5 d x = \frac{x^{4}}{4} + \frac{4 x^{3}}{3} + 5 x + C$ $int x^3 + 4x^2 + 5dx = x^4/4 + (4x^3)/3 + 5x + C$

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How do you evaluate the integral $\int x^{3} + 4 x^{2} + 5 d x$ $intx^3+4x^2+5 dx$ ?

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