Question

In Exercises 31-36, evaluate the integral using the following values. $$\int_{2}^{4} x^{3} d x=60, \quad \int_{2}^{4} x d x=6, \quad \int_{2}^{4} d x=2$$ $$ \int_{2}^{4}\left(6+2 x-x^{3}\right) d x $$

Short answer

-36

Step-by-step solution

Splitting the integral

The integral can be split into a sum of three separate integrals according to the properties of integrals: \[ \int_{2}^{4}(6 + 2x - x^3)dx = \int_{2}^{4} 6dx + \int_{2}^{4} 2xdx - \int_{2}^{4} x^3dx \]. The constant 6 and 2 get factored out from the integral and leave us with \[ 6\int_{2}^{4} dx + 2\int_{2}^{4} xdx - \int_{2}^{4} x^3dx \].

Substitute given integral values

Next, the given values for separate integrals can now be substituted in. Recall that the given values are \[ \int_{2}^{4} x^{3}dx=60, \int_{2}^{4} xdx=6, \int_{2}^{4} dx=2 \].

Compute the final result

Finally, we can compute the total integral as follows: \[6 \int_{2}^{4} dx + 2\int_{2}^{4} xdx - \int_{2}^{4} x^3dx = 6*2+2*6-60 = 12+12-60=-36\].