The integral 2𝜋 4xe−x2 dx can be done with the substitution u = x2 and du =

To solve the integral \(\int e^{(4x-2)} \, dx\), we can use the method of substitution. Let \( u = 4x - 2 \). Then, we need to find \( du \). The derivative of \( u \) with respect to \( x \) is: \[ \frac{du}{dx} = 4 \] So, \[ du = 4 \, dx \] \[ dx = \frac{1}{4} \, du \] Now, substitute \( u \) and \( dx \) into the integral: \[ \int e^{(4x-2)} \, dx = \int e^u \cdot \frac{1}{4} \, du \] \[ = \frac{1}{4} \int e^u \, du \] The integral of \( e^u \) is \( e^u \), so: \[ \frac{1}{4} \int e^u \, du = \frac{1}{4} e^u + c \] Substitute back \( u = 4x - 2 \): \[ \frac{1}{4} e^{(4x-2)} + c \] Therefore, the correct answer is: \[ \boxed{D} \]

Use the Table of Integrals to find x25 − 4x2 dx.

Step 2Now we can say that the volume of the solid created by rotating the region under y = 4e−x2and above the x-axis between x = 0 and x = 1 around the y-axis isV = b2𝜋rh dxa= 1 1 2𝜋x $$4e−x2  dx.0Step 3The integral 2𝜋 4xe−x2 dx can be done with the substitution u = $$−x2 and du = $$−2x dx.Step 4With the substitution, we have 2𝜋 4xe−x2 dx = − $$4π eu du = − $$4πeu + C.Step 5Going back to x, the volume of our solid is

Integrate[xPower[\(40)2x+5\(41),8],x]

Integrate[6xPower[Square[3x]+4,4],x]