Question

1. find $int_{0}^{1}e^{3x + 1}dx$ 5. find $int_{0}^{1}\frac{x^{2}}{x^{3}+1}dx$

Explanation:

1. For $\int_{0}^{1}e^{3x + 1}dx$:

• Step1: Use substitution

Let $u = 3x+1$, then $du=3dx$ and $dx=\frac{1}{3}du$. When $x = 0$, $u = 1$; when $x = 1$, $u=4$. So $\int_{0}^{1}e^{3x + 1}dx=\frac{1}{3}\int_{1}^{4}e^{u}du$.

• Step2: Integrate $e^{u}$

The antiderivative of $e^{u}$ is $e^{u}$. So $\frac{1}{3}\int_{1}^{4}e^{u}du=\frac{1}{3}[e^{u}]_{1}^{4}=\frac{e^{4}-e}{3}$.

1. For $\int_{0}^{1}\frac{x^{2}}{x^{3}+1}dx$:

• Step1: Use substitution

Let $t=x^{3}+1$, then $dt = 3x^{2}dx$ and $x^{2}dx=\frac{1}{3}dt$. When $x = 0$, $t = 1$; when $x = 1$, $t=2$. So $\int_{0}^{1}\frac{x^{2}}{x^{3}+1}dx=\frac{1}{3}\int_{1}^{2}\frac{1}{t}dt$.

• Step2: Integrate $\frac{1}{t}$

The antiderivative of $\frac{1}{t}$ is $\ln|t|$. So $\frac{1}{3}\int_{1}^{2}\frac{1}{t}dt=\frac{1}{3}[\ln(t)]_{1}^{2}=\frac{\ln(2)}{3}$.

Answer:

1. For $\int_{0}^{1}e^{3x + 1}dx$: $\frac{e^{4}-e}{3}$
2. For $\int_{0}^{1}\frac{x^{2}}{x^{3}+1}dx$: $\frac{\ln(2)}{3}$