Calculus definite integrals help #3?

Question

Evaluate the definite integral below

X(3X^2-1)^3dx
upper = 1
lower = 0

Answer 1

Try the substitution rule.
let u = 3x^2 - 1
so du = 6xdx

to complete the substitution, the integrand must contain a 6. To keep the integral the same expression, multiply the integral by 1/6.

1/6 int u^3 du equals 1/6 1/4 (u^4) with upper = 2 and lower = -1, because the limit is now u. these limits are found by the upper and lower limits of x. Put these values into u = 3x^2 - 1.

Do the antiderivative calculation and the answer = 5/8.

Answer 2

Take y=(3x^2 - 1)
then dy/dx = 6x

integral from 0 to 1 x(3x^2 - 1)^3 dx =
= integral from 0 to 1 (3x^2 - 1)^3 * 6x/6 dx =
= integral from -1 to 2 y^3/6 dy =
= y^4/24 from -1 to 2 = (2^4)/24 - (-1)^4 / 24 =
= 16/24 - 1/24 = 15/24 = 5/8

Answer 3

u-sub
u = 3x^2
du = 6x dx

now you have integral (1/6)(u-1)^3 du
factor out the cube

change limits if you want, upper = 3
, lower = 0

new integral: -1/6 + u/2 - (u^2)/2 + (u^3)/6 du
integrate: -u/6 + (u^2)/4 -(u^3)/6 + (u^4)/24
lower of 0 = 0
evalute: -3/6 + 9/4 - 27/6 + 81/24

answer: 5 / 8

Answer 4

a million. confident necessary of algebraic functionality 3 $ 3x^2 + 5x - 4 dx a million bear in mind massive important regulation of integration: confident necessary=F(a)-F(b) and we are gonna do the required of each section via itself $(all that stuff)= x^3 +5/2x^2 - 4x evaluated from a million to 3 so then we evaluate it (3^3 + (5/2)3^2 - 4(3)) - (a million^3 + (5/2)a million^2 - 4(a million)) =27 + forty 5/2 - 12 - a million - 5/2 + 4 3. confident necessary of trig functionality its an analogous component different than the antiderivative of cos is sin pi/2 $ 2t + value dt -pi/2 =t^2 +sint evaluated from unfavorable pi/2 to helpful pi/2 =(pi^2)/4 + a million - ( (pi^2)/4 -a million) advert then the pis cancel and you get subtraction of a unfavorable so all of it equals 2. 4. indefinite necessary $ t^3 (t^4 +5)^a million/2 dt this one is somewhat troublesome in case you do no longer see the substitution u=t^4+5 and du=t^3dt so our new necessary is $u^a million/2du so we try this and get = (2/3)u^3/2 +c we replace u with t^4+5 =(2/3)(t^4+5)^3/2 +c and there you pass the 2d now makes use of an analogous sort of substitution purely with trig $(tanx)^a million/2(sec^2(x))dx u=tanx du=sec^2(x)dx all of us understand that the spinoff of tanx is sec^2x new necessary $u^a million/2du =(2/3)u^3/2 +c replace =(2/3)(tanx)^3/2 +c wish those help some