All Flashcards

Steps to solve $\int xe^x , dx$ using Integration by Parts?

Choose $u=x$ , $dv=e^x dx$ . 2. Find $du=dx$ , $v=e^x$ . 3. Apply formula: $xe^x - \int e^x dx$ . 4. Evaluate: $xe^x - e^x + C$ .

Steps to solve $\int \ln(x) , dx$ using Integration by Parts?

Rewrite as $\int 1 \cdot \ln(x) , dx$ . 2. Choose $u=\ln(x)$ , $dv=dx$ . 3. Find $du=\frac{1}{x}dx$ , $v=x$ . 4. Apply formula: $x\ln(x) - \int x(\frac{1}{x}) dx$ . 5. Evaluate: $x\ln(x) - x + C$ .

Steps to solve $\int x^2 \cos(x) , dx$ ?

Choose $u=x^2$ , $dv=\cos(x)dx$ . 2. Find $du=2x dx$ , $v=\sin(x)$ . 3. Apply formula: $x^2\sin(x) - \int 2x\sin(x) dx$ . 4. Apply IBP again to $\int 2x\sin(x) dx$ . 5. Final result: $x^2 \sin(x) + 2x \cos(x) - 2\sin(x) + C$ .

Steps to evaluate $\int e^x \cos x , dx$ ?

Apply IBP twice, keeping $e^x$ as part of 'u' or 'dv' consistently. 2. Isolate the original integral using algebraic manipulation. 3. Solve for the original integral.

Explain the purpose of Integration by Parts.

To simplify integrals involving the product of two functions by transforming them into simpler integrals.

How does Integration by Parts relate to the product rule?

It's the reverse process of the product rule for differentiation.

Why is choosing 'u' and 'dv' important?

Proper selection simplifies the integral, making it easier to solve.

What are the differences between u-substitution and Integration by Parts?

u-substitution: Reverses the chain rule, simplifies composite functions. | Integration by Parts: Reverses the product rule, simplifies products of functions.