Question

instructions

1. work out this problem on one (1) sheet of plain paper. label is fe work out #2.
2. scan this sheet together with the other workout questions and submit in canvas, in the final exam workout questions portal.

work - out #2
a student determines that \\(\frac{d}{dx}\int_{1}^{\sin x} 4t^{3}dt = 4\sin^{3}x\\) by applying the fundamental theorem of calculus.
a. (5 pts) identify the error in the students work. write your response in no more than two (2) grammatically correct, complete sentences.
b. (5 pts) solve the problem correctly.

Explanation:

Part A

The student forgot to apply the chain rule. When differentiating a definite integral with a function (here $\sin x$) as the upper limit, we must multiply by the derivative of the upper limit function. So, they should have included the derivative of $\sin x$ (which is $\cos x$) in their calculation.

Step 1: Recall the Fundamental Theorem of Calculus (FTC) and Chain Rule

The FTC for differentiation of an integral with a function as the upper limit states that if $F(x)=\int_{a}^{u(x)}f(t)dt$, then $F^\prime(x)=f(u(x))\cdot u^\prime(x)$. Here, $f(t) = 4t^{3}$, $u(x)=\sin x$, and $a = 1$.

Step 2: Apply the formula

First, find $f(u(x))$: substitute $t=\sin x$ into $f(t)$, so $f(\sin x)=4(\sin x)^{3}$. Then, find $u^\prime(x)$: the derivative of $\sin x$ with respect to $x$ is $\cos x$.
Now, multiply these two results according to the chain - rule version of the FTC: $\frac{d}{dx}\int_{1}^{\sin x}4t^{3}dt=4(\sin x)^{3}\cdot\cos x = 4\sin^{3}x\cos x$.

Answer:

The student's error is not applying the chain rule; they failed to multiply the result by the derivative of the upper - limit function $\sin x$ (i.e., $\cos x$) when using the Fundamental Theorem of Calculus.

Part B