What is the definite integral of 2x^2 + 4x + 1 with a lower limit of 3 and a higher limit of 6?

When integrating we add one to the power of x and divide the number in front of the x by the new power for each part of the function. So 2x^2 becomes 2/3 x^3, 4x becomes 4/2 x^2, 1 becomes 1/1 x^1. Then we need to put in the values of the upper and lower limits for x. So put the upper limit of 6 in first then subtract the value of putting the lower limit of 3 in. [(2/3 * 6^3) + (2 * 6^2) + (6)] - [ (2/3 * 3^3) + (2 * 3^2) + (3)] = (222) -(39) = 183 and that is the final answer.

Answered by Ed L. Maths tutor

2918 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

In a science experiment a substance is decaying exponentially. Its mass, M grams, at time t minutes is given by M= 300e^-0. 5t


Solve the equation: log5 (4x+3)−log5 (x−1)=2.


If f(x)=x^2 and g(x)=5x-11, then what is fgg(x) when x=3?


Prove the identity (4cos(2x))/(1+cos(2x)) = 4-2sec^2(x)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences