ediculeN

2021-02-12

Solve. $\int \left(4a{x}^{3}+3b{x}^{2}+2cx\right)dx$

ottcomn

There are three properties of integrals that we'll exploit to evaluate this. Those are
1.The integral of a sum is the sum of the integrals of the terms.
$\int \left[f\left(x\right)+g\left(x\right)\right]dx=\int f\left(x\right)dx+\int g\left(x\right)dx$
2.The integral of a constant times a function is equal to the constant times the integral of the function.
$\int kf\left(x\right)dx=k\int f\left(x\right)dx$
3.The integral of a power n of x is given by
$\int {x}^{n}dx=\left({x}^{\left(}n+1\right)\right)/\left(n+1\right)+C$
Now let's use those. Firstly, we use property 1 to split up that integral.
$\int \left(4a{x}^{3}+3b{x}^{2}+2cx\right)dx=\int 4a{x}^{3}dx+\int 3b{x}^{2}dx+\int 2cxdx$
Next we'll use property 2 to pull out any constants.
$4a\int {x}^{3}dx+3b\int {x}^{2}dx+2c\int xdx$
And finally we'll use property 3 to evaluate those integrals.
$=4a\left({x}^{4}\right)/4+3b\left({x}^{3}\right)/3+2c\left({x}^{2}\right)/2+C$
Now we just simplify and we're done!
$a{x}^{4}+b{x}^{3}+c{x}^{2}+C$

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