# Application of 2nd fundamental theorem of calculus

I would like to clarify the usage of the 2nd fundamental theorem of calculus, in 3 parts. For all 3 parts, consider $F'(a)$ to be

$$F'(a) = \frac{1}{1+a+a^2}$$

The questions are:

1. Find $\frac{d}{dy}\int^y_1 F'(a) \,da$
2. Find $\frac{d}{dy}\int^1_y F'(a) \,da$
3. Find $\frac{d}{dy}\int^{y^2}_1 F'(a)\, da$

My working for each of the questions are:

1.

\begin{align*} \frac{d}{dy} \int^y_1 F'(a) \,da &= \frac{d}{dy}\begin{bmatrix}F(y) - F(1)\end{bmatrix} \\&=F'(y) \\&=\frac{1}{1+y+y^2}. \end{align*}

2.

\begin{align*} \frac{d}{dy} \int^1_y F'(a) \,da &= \frac{d}{dy}\begin{bmatrix}F(1) - F(y)\end{bmatrix} \\&=-F'(y) \\&=\frac{-1}{1+y+y^2}. \end{align*}

3.

\begin{align*} \frac{d}{dy} \int^{y^2}_1 F'(a) \,da &= \frac{d}{dy}\begin{bmatrix}F(y^2) - F(1)\end{bmatrix} \\&=F'(y^2) \\&=\frac{2y}{1+y^2+y^4}. \end{align*}

UPDATE: I shall update (3) with the chain rule working.

\begin{align*} \frac{d}{dy} \int^{y^2}_1 F'(a) \,da &= \frac{d}{dy}\begin{bmatrix}F(y^2) - F(1)\end{bmatrix} \\&=F'(y^2)(2y) \\&=\frac{1}{1+(y^2)+(y^2)^2} (2y) \tag{by chain rule} \\&=\frac{2y}{1+y^2+y^4}. \end{align*}

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The third one would need a chain rule. Though, you do have used it for final answer. – hjpotter92 Mar 8 '13 at 12:59

Your answers are correct. However, one should note that$\dfrac{dF(y^2)}{dy}=F'(y^2)\cdot2y$(Chain rule). Your final answers are perfectly fine but the intermediate step is wrong.