How do you evaluate the definite integral #int (x^2-x+6)dx# from [1,4]? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer James May 21, 2018 #color(red)[int_1^4(x^2-x+6)*dx=63/2]# Explanation: show below #color(red)[int_1^4(x^2-x+6)*dx=]# #[1/3*x^3-1/2*x^2+6x]_1^4=# #[(64/3-16/2+24)-(1/3-1/2+6)]=[63/2]# 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 2200 views around the world You can reuse this answer Creative Commons License