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

The Fundamental Theorem of Calculus - Part 2

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Fundamental Theorem of Calculus (FTC) - Part 2

Computing definite integrals by definition is often difficult and time consuming. Fortunately, the second part of The Fundamental Theorem of Calculus provides us with a straightforward way of computing definite integrals.

Fundamental Theorem of Calculus: 2

Let f(x)f(x) be a continuous function on [a,b][a,b], so that F′(x)=f(x)F'(x)=f(x) on [a,b][a,b]we have
∫abf(x)dx=F(b)−F(a)\boxed{\int^b_af(x)dx=F(b)-F(a)}

Note: Sometimes we will use the notation F(x)∣ab=F(b)−F(a)F(x) \bigg|_a^b=F(b)-F(a)

Wize Concept
To compute definite integrals using FTC, compute the antiderivative of the function, substitute the upper bound, and subtract substituting the lower bound.

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Example: FTC Part 2

Evaluate the integral using the Fundamental Theorem of Calculus.

∫123xdx\displaystyle \int_1^23\sqrt{x}dx

∫123xdx\displaystyle \int_1^23\sqrt{x}dx

=∫123x12dx \displaystyle=\int_1^23x^{\frac{1}{2}}dx\

=3⋅x3232∣12=\displaystyle \left. \displaystyle 3\cdotp \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right|_1^2 (by defining the antiderivative)


=2⋅(2)32−2(1)32=\displaystyle2 \cdotp (2)^{\frac{3}{2}}-2(1)^{\frac{3}{2}} (applying FTC2)

=42−2=4\sqrt{2}-2

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Example: FTC Part 2

Evaluate the integral using the Fundamental Theorem of Calculus.

∫14e−3x+2x−1x12dx\displaystyle \int_1^4e^{-3x}+2^x-\frac{1}{x^{\frac{1}{2}}}dx


∫14e−3x+2x−1x12dx\displaystyle \int_1^4e^{-3x}+2^x-\frac{1}{x^{\frac{1}{2}}}dx

=∫14(e−3x+2x−x−12)dx=\displaystyle \int_1^4 (e^{-3x}+2^x-{x^{-\frac{1}{2}}})dx

=[−13e−3x+1ln⁡22x−(1)x1212]14=\displaystyle \left[-\frac{1}{3}e^{-3x}+\frac{1}{\ln2}2^x-\left(1\right)\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]_1^4

=[−13e−3(4)+1ln⁡224−(2)412]−[−13e−3(1)+1ln⁡221−(2)112]=−13e12+16ln⁡2−4−[−13e3+2ln⁡2−2]=−13e12+13e3+14ln⁡2−2\begin{aligned} &= \left[ -\frac{1}{3}e^{-3(4)} + \frac{1}{\ln2}2^4 - (2)4^{\frac{1}{2}}\right] - \left[ -\frac{1}{3}e^{-3(1)} + \frac{1}{\ln2}2^1 - (2)1^{\frac{1}{2}}\right]\\ \\ &= \frac{-1}{3e^{12}} + \frac{16}{\ln2}-4-\left[ \frac{-1}{3e^3} + \frac{2}{\ln2} -2\right]\\ \\ &=-\frac{1}{3e^{12}}+\frac{1}{3e^3}+\frac{14}{\ln2}-2 \end{aligned}

Evaluate ∫−20∣x2−1∣dx\displaystyle \int_{-2}^0\left|x^2-1\right|dx.
If ∫02[3(kx)2+(k+1)x+1]dx=10\displaystyle \int_0^2[3(kx)^2+(k+1)x+1]dx=10, then what is a possible value of kk ?
Given that both f(x)f\left(x\right) andf′(x)f'\left(x\right) are continuous everywhere, iff(1)=10f\left(1\right)=10 and ∫−31f′(x)dx=12\displaystyle \int_{-3}^1f'\left(x\right)dx=12, find f(−3)f\left(-3\right).