# Use the Fundamental Theorem of Calculus to find G'(x) if: \\ G(x)=\int_1^{x^2} \cos t dt

## Question:

Use the Fundamental Theorem of Calculus to find {eq}G'(x){/eq} if:

{eq}G(x)=\int_1^{x^2} \cos t dt{/eq}

## Fundamental Theorem of Calculus:

The fundamental theorem of calculus is very helpful in finding the derivative of a definite integral without actually computing the definite integral. This theorem states:

$$\frac{d}{d x} \int_{a}^{x} f(t) d t=f(x)$$

The given integral is:

$$G(x)=\int_1^{x^2} \cos t dt$$

To apply the fundamental theorem of calculus for finding its derivative, the upper limit of the integral should be a variable of degree 1.

We make this by substitution.

\begin{align} &\text{Let } x^2 =t \\ &\text{Then } 2x\, dx = dt \\ \\ &\text{Limits:} \\ &\text{Upper limit: }: t=x^2 \\ &\text{Lower limit: } t=1 \Rightarrow x^2=1 \Rightarrow x=1 \end{align}

Substitute these values in the given integral:

$$G(x) = \int_1^t \cos (x^2) \, (2x \, dx) = \int_1^t 2x \cos(x^2) dx$$

Now we interchange the variables x and t in the above integral:

$$G(x) = \int_1^x 2t \cos(t^2) dt$$

Now we use the fundamental theorem of calculus to find its derivative which states:

$$\dfrac{d}{d x} \int_{a}^{x} f(t) d t=f(x)$$

Then we get:

$$G'(x) =\dfrac{d}{d x} \int_1^x 2t \cos(t^2) dt = 2x \cos(x^2) dx$$

Therefore: {eq}\boxed{\mathbf{G'(x)= 2x \cos(x^2) dx}} {/eq}