Question

His question is designed to be answered without a calculator. Two students wrote antiderivative formulas for k sin (kx), where k > 1. Student 1: Integral of (k sin (k x) ) d x = k squared cosine (k x) + C Student 2: Integral of (k sine (k x) ) d x = negative cosine (k x) + C Which student, if any, wrote a correct antiderivative formula? Student 1 only Student 2 only both Student 1 and Student 2 neither Student 1 nor Student 2

Answer 1

Student 2 only

Explanation:

Given that:

The written antiderivatives formulas by the students for k sin (kx), where k > 1 are:

Student 1 :
`\int (k \ sin (kx)) \ dx= k^2 \ cos (kx) + C`

Student 2:
`\int (k \ sin (kx) ) \ dx = - cos (kx) + C`

The student that wrote the correct anti-derivate formula is Student 2 only.

This is because:

Suppose:

`I = \int (k \sin (kx) ) \ dx` where; k > 1

This implies that:

`I = \int (k \sin (kx) ) \ dx`

`I =k \bigg ( (-cos (kx))/(k) \bigg) + C` because
`\int sin (ax) \ dx = (-cos (ax))/(a)+C`

here;

C = constant of the integration

∴

I = - cos (kx) + C which aligns with the answer given by student 2 only.