Introduction to Integrals

Problem : Evaluate x(2x ²+4)³ dx .

Solution for Problem 1 >>

u		= 2x 2 +4
		= 4x
dx		=

Now substitute back into the expression:

x u 3		= u 3 du =
		= 2x 2+4 + c

Close

Problem : Evaluate 4x cos(x ²+2)dx.

Solution for Problem 2 >>

u		= x 2 +2
		= 2x
dx		=
4x cos(u)		= 2cos(u)du = 2 sin(u)
		= 2 sin(x 2 + 2) + c

Close

Problem : Evaluate (x )dx.

Solution for Problem 3 >>

u		= x 2 -5
		= 2x
dx		=
x		= du = = u
		= x 2-5 + c

Close

Problem : Evaluate (3x-4)² dx.

Solution for Problem 4 >>

		u = 3x - 4
		= 3
		dx = du
		u 2 du

(Note: the limits of integration have been removed, since they do not apply to u but to x .

u 2 du = u 3

Now substitute x back into the result.

		= 3x-4 0 1
		= 3(1)-4-(-4)3
		= (-1)-(-64)
		= = 7

Close

Problem : Use the trapezoid rule with five subdivisions to approximate the area under f (x) = x ² + 1 on [1, 6] .

Solution for Problem 5 >>

Area		2+2(5)+2(10+2(17)+2(26)+37(1)
		= 77.5

This compares well with

16x2+1 dx=75

Close