Solve ∫ (x^2-1)e^-02x

$$\int ( x ^ { 2 } - 1 ) e ^ { - 02 x }$$

Evaluate

$$\frac{x^{3}}{3}-x+С$$

Solution Steps

Multiply 0 and 2 to get 0.

$$\int \left(x^{2}-1\right)e^{0x}\mathrm{d}x$$

Anything times zero gives zero.

$$\int \left(x^{2}-1\right)e^{0}\mathrm{d}x$$

Calculate e to the power of 0 and get 1.

$$\int \left(x^{2}-1\right)\times 1\mathrm{d}x$$

Use the distributive property to multiply x^{2}-1 by 1.

$$\int x^{2}-1\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int -1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}.

$$\frac{x^{3}}{3}+\int -1\mathrm{d}x$$

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}-x$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{x^{3}}{3}-x+С$$

Differentiate w.r.t. x

$$x^{2}-1$$