John Abbott 201-SN3-RE tutoring

For A, look for the derivative of the exponent. For B, ask which factor becomes simpler when differentiated.

1. A: use the chain rule in reverse

   The derivative of $x^2$ is $2x$, exactly the factor multiplying the exponential. Set $u=x^2$, where u is the new integration variable. Then $du=2x\,dx$. The x-bounds 0 and 1 become u-bounds 0 and 1.

   $$\begin{aligned}A&=\int_0^1 e^u\,du\\&=[e^u]_0^1\\&=e-1.\end{aligned}$$

   The bracket means the value at the upper bound minus the value at the lower bound. The lower contribution is $e^0=1$.

2. B: simplify the product with integration by parts

   Differentiating x gives 1, while $e^x$ is easy to integrate. Choose $u=x$ and $dv=e^x\,dx$, so $du=dx$ and $v=e^x$. Here, u is the factor we differentiate, and v is an antiderivative of $e^x$.

   Integration by parts reverses the product rule: $$\int u\,dv=uv-\int v\,du.$$ Thus $$\begin{aligned}B&=[xe^x]_0^1\\&\quad-\int_0^1 e^x\,dx\\&=e-(e-1)=1.\end{aligned}$$

   A product alone does not decide the method. In A, the factor $2x$ matches the derivative of the inner function $x^2$. In B, differentiating x makes the remaining integral simpler.

3. Check by differentiating

   For A, differentiating $e^{x^2}$ gives $2xe^{x^2}$. For B, an antiderivative is $(x-1)e^x$: differentiating it with the product rule gives $$e^x+(x-1)e^x=xe^x.$$ At $x=1$ it is 0; at $x=0$ it is −1. The definite integral is $0-(-1)=1$.