How do you evaluate int 2x^3-7 for [2,5]?

Oct 10, 2015

This is basically finding f(x) with the diffrential given.

Explanation:

So integration, (and points)

I'll just ignore typing the integral sign.

So first split the equation.
Int. of 2x^3 - Int. of 7
power of x increases +1 and becomes denominator, so.

x^4/2 - 7x

y=x^4/2 - 7x + c

Insert 2,5
5=(2)^4/2 - 7(2) + c
5= -6+c
11=c

=> y=(x^4)/2-7x+11

Oct 11, 2015

${\int}_{2}^{5} \left(2 {x}^{3} - 7\right) \mathrm{dx}$

Explanation:

${\int}_{2}^{5} \left(2 {x}^{3} - 7\right) \mathrm{dx}$ by the Fundamental Theorem of Calculus.
(If you need to do this by using the definiion of definite integral, message me, or comment, or repost the question with "by using the definition".)

To use the Fundamental Theorem of Calculus.

Find an anti derivative of $2 {x}^{2} - 7$.

An antiderivative is $\frac{2 {x}^{4}}{4} - 7 x = {x}^{4} / 2 - 7 x$

Now, evaluate the antiderivative from $2$ to $5$:

${\int}_{2}^{5} \left(2 {x}^{3} - 7\right) \mathrm{dx} = {\left({x}^{4} / 2 - 7 x\right]}_{2}^{5}$

$= \left({\left(5\right)}^{4} / 2 - 7 \left(5\right)\right) - \left({\left(2\right)}^{4} / 2 - 7 \left(2\right)\right)$

$= \frac{567}{2}$