# Find f. f '(t) = 20/1 + t^{2} f(1) = 0

**Solution:**

Integrals are the values of the function found by the process of integration.

The process of getting f(x) from f'(x) is called integration.

Given f '(t) = 20/ 1 + t^{2} and f(1) = 0

We know that by definition of __integral__ of tan⁻¹(x) = 1/1 + t^{2}

f '(t) = 20 / 1 + t^{2} = 20(1/ 1 + t^{2})

__Integrate__ on both sides, we get

f(t) = 20tan⁻¹(t) + c

f(1) = 0; f(1) = 20tan⁻¹(1) + c = 0

c = -20(π/4) = -5π.

Therefore, f(x) = 20tan⁻¹(t) -5π.

## Find f. f '(t) = 20/1 + t^{2} f(1) = 0

**Summary:**

If f '(t) =20/1 + t^{2} f(1) = 0, then f is f(x) = 20tan⁻¹(t) - 5π.

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