To find the antiderivative [tex]\( F(t) \)[/tex] of the given function [tex]\( f(t) = 2 \sec^2(t) - 5t^3 \)[/tex] and ensure that [tex]\( F(0) = 0 \)[/tex], we follow these steps:
1. Identify the components: The function we need to integrate is [tex]\( f(t) = 2 \sec^2(t) - 5 t^3 \)[/tex].
2. Integrate each term separately:
- First term: Integrate [tex]\( 2 \sec^2(t) \)[/tex].
The integral of [tex]\( \sec^2(t) \)[/tex] is [tex]\(\tan(t)\)[/tex]. Therefore,
[tex]\[
\int 2 \sec^2(t) \, dt = 2 \tan(t).
\][/tex]
- Second term: Integrate [tex]\(-5t^3\)[/tex].
The integral of [tex]\( t^3 \)[/tex] is [tex]\( \frac{t^4}{4} \)[/tex]. Therefore,
[tex]\[
\int -5 t^3 \, dt = -5 \int t^3 \, dt = -5 \cdot \frac{t^4}{4} = -\frac{5t^4}{4}.
\][/tex]
3. Combine the results:
Adding the antiderivatives of each term gives us the general form of [tex]\( F(t) \)[/tex]:
[tex]\[
F(t) = 2 \tan(t) - \frac{5t^4}{4} + C,
\][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
4. Determine the constant [tex]\( C \)[/tex]:
We apply the condition [tex]\( F(0) = 0 \)[/tex]:
[tex]\[
F(0) = 2 \tan(0) - \frac{5 \cdot 0^4}{4} + C = 0.
\][/tex]
Since [tex]\(\tan(0) = 0\)[/tex], we get
[tex]\[
0 + 0 + C = 0 \implies C = 0.
\][/tex]
5. Final antiderivative:
Substitute [tex]\( C = 0 \)[/tex] back into the antiderivative:
[tex]\[
F(t) = 2 \tan(t) - \frac{5t^4}{4}.
\][/tex]
Therefore, the antiderivative [tex]\( F(t) \)[/tex] of the function [tex]\( f(t) = 2 \sec^2(t) - 5 t^3 \)[/tex] with [tex]\( F(0) = 0 \)[/tex] is:
[tex]\[
F(t) = -\frac{5t^4}{4} + 2 \frac{\sin(t)}{\cos(t)}.
\][/tex]
So,
[tex]\[
F(t) = -\frac{5 t^4}{4} + 2 \tan(t).
\][/tex]
