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**Show that the operator A defined by A[y](t) = integ(s(y(s))ds) [integration from t to a] is linear. (Hint: Use integration by parts.)**

**Attempt at the solution:**

A[y](t) = integ(s(y(s))ds) [integ from t to a]

Using integration by parts:

Let u = s

du = ds

let dv = y(s)

v = 1/2 y^2(s)

so,

A[y](t) = s(1/2y^2(s)) - integ(ds 1/2y^2(s)) [i'm not sure how to change the integration limits]

A[y](t) = 1/2s y^2(s) - [1/2 (1/3 y^3(s))] (from ___ to ___)

and just sub in the integrating limits and simplify?

am i doing something wrong.

A[y](t) = integ(s(y(s))ds) [integ from t to a]

Using integration by parts:

Let u = s

du = ds

let dv = y(s)

v = 1/2 y^2(s)

so,

A[y](t) = s(1/2y^2(s)) - integ(ds 1/2y^2(s)) [i'm not sure how to change the integration limits]

A[y](t) = 1/2s y^2(s) - [1/2 (1/3 y^3(s))] (from ___ to ___)

and just sub in the integrating limits and simplify?

am i doing something wrong.