Question

evaluate the integral \\(\int 30t^4 e^{-t^5} dt\\). \\(\int 30t^4 e^{-t^5} dt = \square\\)

Explanation:

Step1: Choose substitution

Let \( u = -t^5 \). Then, find \( du \):
Differentiate \( u \) with respect to \( t \): \( \frac{du}{dt} = -5t^4 \), so \( du = -5t^4 dt \), and \( -6 du = 30t^4 dt \) (since \( 30t^4 dt = -6 \times (-5t^4 dt) \)).

Step2: Substitute into integral

The integral \( \int 30t^4 e^{-t^5} dt \) becomes \( \int e^{u} \times (-6 du) \)? Wait, no, wait. Wait, \( u = -t^5 \), so \( e^{-t^5} = e^u \), and \( 30t^4 dt \): from \( du = -5t^4 dt \), we can solve for \( t^4 dt = \frac{du}{-5} \), so \( 30t^4 dt = 30 \times \frac{du}{-5} = -6 du \)? Wait, that seems off. Wait, no, let's do it again. Let \( u = -t^5 \), then \( du/dt = -5t^4 \), so \( dt = du / (-5t^4) \). Then substitute into the integral:

\( \int 30t^4 e^{u} \times \frac{du}{-5t^4} \). The \( t^4 \) terms cancel. Then \( 30 / (-5) = -6 \), so the integral becomes \( \int -6 e^u du \). Wait, but that would be \( -6 e^u + C \), but let's check the derivative. Wait, maybe I made a mistake in the substitution. Wait, alternatively, let's see: the derivative of \( -t^5 \) is \( -5t^4 \), so \( 30t^4 dt = -6 \times (-5t^4 dt) = -6 du \). So \( \int 30t^4 e^{-t^5} dt = \int e^{u} \times (-6 du) \)? No, wait, \( e^{-t^5} = e^u \), and \( 30t^4 dt = -6 \times (-5t^4 dt) = -6 du \), so the integral is \( \int e^u \times (-6 du) = -6 \int e^u du = -6 e^u + C \). But let's check by differentiating the result. Let's take \( F(t) = -6 e^{-t^5} \), then \( F'(t) = -6 \times e^{-t^5} \times (-5t^4) = 30t^4 e^{-t^5} \), which matches the integrand. So that works. Wait, so the integral is \( -6 e^{-t^5} + C \)? Wait, but let's do the substitution again. Let \( u = -t^5 \), then \( du = -5t^4 dt \), so \( 30t^4 dt = -6 du \) (since \( 30t^4 dt = 30 \times (dt) \times t^4 \), and \( dt = du / (-5t^4) \), so \( 30t^4 \times (du / (-5t^4)) = -6 du \)). Then the integral \( \int 30t^4 e^{-t^5} dt = \int e^u \times (-6 du) = -6 \int e^u du = -6 e^u + C = -6 e^{-t^5} + C \). Let's verify by differentiating: \( d/dt [ -6 e^{-t^5} ] = -6 \times e^{-t^5} \times (-5t^4) = 30t^4 e^{-t^5} \), which is the integrand. So that's correct.

Wait, but maybe I messed up the sign. Let's check again. Let \( u = -t^5 \), so \( du = -5t^4 dt \), so \( 5t^4 dt = -du \), so \( 30t^4 dt = 6 \times 5t^4 dt = 6 \times (-du) = -6 du \). So the integral is \( \int 30t^4 e^u dt = \int e^u \times (-6 du) = -6 e^u + C = -6 e^{-t^5} + C \). Yes, that's correct.

Alternatively, maybe the substitution is \( u = t^5 \), but no, because the exponent is \( -t^5 \). So the correct substitution is \( u = -t^5 \), leading to the integral being \( -6 e^{-t^5} + C \), or simplified, \( -6e^{-t^5} + C \).

Wait, but let's do it step by step again.

Let \( u = -t^5 \).

Then, \( du/dt = -5t^4 \implies dt = du / (-5t^4) \).

Substitute into the integral:

\( \int 30t^4 e^{u} \cdot \frac{du}{-5t^4} \)

The \( t^4 \) terms cancel:

\( \int 30 \cdot \frac{e^u}{-5} du \)

Simplify \( 30 / (-5) = -6 \):

\( \int -6 e^u du \)

Integrate \( e^u \):

\( -6 e^u + C \)

Substitute back \( u = -t^5 \):

\( -6 e^{-t^5} + C \)

Verify by differentiation:

\( d/dt [ -6 e^{-t^5} ] = -6 \cdot e^{-t^5} \cdot (-5t^4) = 30t^4 e^{-t^5} \), which matches the integrand. So the antiderivative is \( -6e^{-t^5} + C \).

Answer:

\( -6e^{-t^5} + C \) (or equivalently \( -6 e^{-t^5} + C \))