The method known as Integration by Parts is a powerful tool in calculus, primarily utilized when dealing with the integration of functions that are products of one another. Derived from the product rule of differentiation, this technique simplifies the integral of a product into an integral that is significantly more manageable to solve. This approach is particularly beneficial when the integral is not straightforward to compute.
| Topic | Problem | Solution |
|---|---|---|
| None | Evaluate. (Be sure to check by differentiating!) … | Given the integral \(\int(4 t^{3}-7) t^{2} dt\), we need to evaluate it. |
| None | Evaluate. (Assume $x>0$.) Check by differentiatin… | Identify the parts of the integral that will be 'u' and 'dv'. In this case, let 'u' be \(\ln x\) an… |
| None | Evaluate the indefinite integral. \[ \int 2 e^{2 … | First, we recognize that this integral is a good candidate for integration by parts, which is given… |
| None | Question 1. Graphical Integration Challenge: u-Su… | Let $F(x)=\int_{-1}^{x} f(t) \mathrm{d} t$ and $G(x)=\int_{0}^{x} g(t) \mathrm{d} t$, where the gra… |
| None | Question 1. Graphical Integration Challenge: $u$-… | Given that $F(x)=\int_{-1}^{x} f(t) \mathrm{d} t$ and $G(x)=\int_{0}^{x} g(t) \mathrm{d} t$, we are… |
| None | Use the substitution $u=-x$ to evaluate $\int_{-2… | Let $u = -x$. Then, $x = -u$ and $d x = -d u$. Also, when $x = -2$, $u = 2$, and when $x = 2$, $u =… |
| None | Using integration by parts find $\int x \sin x \m… | Choose u = x and dv = \(\sin x \mathrm{~d} x\), then find du and v: du = \(\mathrm{d} x\), v = -\(\… |
| None | \( \int x^{3} e^{x^{2}} d x \) | \( u = x^{2} \) |
| None | a) \( \int \frac{\operatorname{arctg} x}{1+x^{2}}… | Let \( y = \operatorname{arctg} x \) |