Evaluate the definite integral

∫(0,2) (x-1)^25 dx..

thats how i got stuck

u=x-1, then du=dx
=∫(0,2) u^25du
=(1/26)u^26. i don't know what to do with integral (2,0)..

just keep working away. You have the correct answer

1/26 (x-1)^26 [0,2]
= 1/26 [(1)^26 - (-1)^26]
= 1/26 [1 - 1]
= 0

you have to break up your integral into two parts, since there is an x-intercept in your domain from 0 to 2, namely x = 1

think of it as finding the area from x=0 to x=1 plus the area from x=1 to x=2

we get ∫-(x-1)^25 dx from 0 to 1
= (-1/26)(x-1)^26 | from 0 to 1
= 0 - (-1/26)(-1)^26 = 1/26
and
∫(x-1)^25 dx from 1 to 2
= (1/26)(x-1)^26 | from 1 to 2
= (1/26)(1)^26 - (1/26)(0) = 1/26

so the total integral is 2/26 or 1/13

To evaluate the definite integral ∫(0,2) (x-1)^25 dx using the substitution u = x-1 and du = dx, you correctly simplified the integral to ∫(0,2) u^25 du.

Now, to evaluate the definite integral from 0 to 2, you need to substitute the limits of integration using the variable u.

When x = 0, u = 0 - 1 = -1.
When x = 2, u = 2 - 1 = 1.

So, you need to rewrite the definite integral in terms of u as follows:

∫(0,2) (x-1)^25 dx = ∫(-1,1) u^25 du.

Now, to find the antiderivative of u^25, you can use the power rule of integration. The power rule states that the integral of x^n with respect to x is (1/(n+1))x^(n+1).

Applying the power rule to u^25, you get:

(1/26)u^26.

Now, substitute the limits of integration:

∫(-1,1) u^25 du = [(1/26)u^26] evaluated from -1 to 1.

Plug in the values of u:

[(1/26)(1^26)] - [(1/26)(-1^26)]

Simplifying:

(1/26)(1) - (1/26)(1) = 1/26 - 1/26 = 0.

Therefore, the value of the definite integral ∫(0,2) (x-1)^25 dx is 0.

To evaluate the definite integral ∫(0,2) (x-1)^25 dx, you started by making a substitution: u = x-1, which gives you du = dx.

Next, you substituted (x-1) with u in the integral to get ∫(0,2) u^25 du.

To evaluate the integral, you can use the power rule of integration, which states that ∫ x^n dx = (1/(n+1)) * x^(n+1) + C, where C is the constant of integration.

Applying the power rule, you can evaluate the integral ∫ u^25 du, which becomes (1/26) * u^26 + C.

Now, you need to substitute back u for x-1. Since the integration limits are (0,2), you need to evaluate the expression at u = 2 and u = 0.

Substituting, you get:

[(1/26) * (2-1)^26] - [(1/26) * (0-1)^26]

Simplifying further:

[(1/26) * 1^26] - [(1/26) * (-1)^26]
= (1/26) * 1 - (1/26) * 1
= 1/26 - 1/26
= 0

Hence, the value of the definite integral ∫(0,2) (x-1)^25 dx is 0.