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Wize AP Calculus (AB) Textbook > Integration & Accumulation of Change

The Fundamental Theorem of Calculus - Part 1

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Fundamental Theorem of Calculus (FTC) - Part I

The first part of The Fundamental Theorem of Calculus explores the direct relationship between derivatives and integrals.

Fundamental Theorem of Calculus: 1

Let f(x)f(x) be a continuous function. Define
F(x)=∫cxf(t)dt\boxed{\quad F(x)=\int_{c}^{x}f(t)\text{d}t\quad }

Then F(x)F(x) is differentiable and it is the antiderivative of f(x)f(x) , that is, F′(x)=f(x)F^{\prime}(x)=f(x).

ddx∫cxf(t) dt=f(x)\boxed{\quad \displaystyle\frac{d}{dx}\int_{c}^{x}f(t)\,\text{d}t=f(x)\quad }

Wize Concept
This means that integration and differentiation are "inverse" operations: if we apply one and then the other, the original function remains unchanged (up to addition of a constant).

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FTC Part 1 & the Chain Rule

If the bounds are functions of xx, we have (by the Chain Rule):

ddx∫cg(x)f(t) dt=f(g(x))⋅g′(x)\boxed{\quad \displaystyle\frac{d}{dx}\int_{c}^{g(x)}f(t)\,\text{d}t=f(g(x)) \cdot g'(x)\quad }

ddx∫h(x)g(x)f(t) dt=f(g(x))⋅g′(x)−f(h(x))⋅h′(x)\boxed{\quad \displaystyle\frac{d}{dx}\int_{h(x)}^{g(x)}f(t)\,\text{d}t=f(g(x)) \cdot g'(x)-f(h(x)) \cdot h'(x)\quad }

Wize Tip
When differentiating a definite integral, there is no need to antiderive the function! Simply substitute the bounds for tt and multiply by the derivatives of the bounds if necessary.

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Example: FTC Part 1

Find the derivative using the Fundamental Theorem of Calculus.

ddx[∫x23t−1t5+1dt]\displaystyle \frac{d}{dx}\left[\int_x^2\frac{3t-1}{t^{5}+1}dt\right]


=−ddx∫2x3t−1t5+1dt=\displaystyle -\frac{d}{dx}\int_2^x\frac{3t-1}{t^5+1}dt (adjust integration limits using the property of definite integral before applying FTC1)

=−3x−1x5+1=-\dfrac{3x-1}{x^5+1} (by FTC 1)
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Example: FTC Part 1

Find ddx[∫3π(t4−1)dt]\displaystyle\frac{d}{dx}\left[\int_3^\pi\left(t^4-1\right)dt\right]

The bounds are real numbers, not functions of xx. Therefore the integral will be evaluated to a number (a constant). But derivative of a constant is zero, so we have:

ddx∫3π(t4−1)dt=0\displaystyle\frac{d}{dx}\int_3^\pi\left(t^4-1\right)dt=0
Find the derivative ddx\displaystyle \frac{d}{dx} of the integral ∫0x2sin⁡t2 dt\displaystyle \int_{0}^{x^2}\sin{t^2}\,\text{d}t.

Compute f(π/3)f(\pi/3)given that f(x)=ddx∫sin⁡xcos⁡xln⁡t dt\displaystyle f(x)=\frac{d}{dx}\int_{\sin x}^{\cos x}\ln t\ dt .