How do you evaluate the definite integral #int root3(t)-2dt# from [-1,1]? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Anjali G Apr 1, 2017 #int_(-1)^(1)(root3(t)-2 )dt# Rewritten: #=int_(-1)^(1)(t^(1/3)-2)dt# #=[3/4t^(4/3)-2t]_(-1)^1# #=(3/4(1)^(4/3)-2(1))-(3/4(-1)^(4/3)-2(-1))# #=color(red)(3/4)-2color(red)(-3/4)-2# #=-4# Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 1513 views around the world You can reuse this answer Creative Commons License