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Wize University Calculus 2 Textbook > Integration

Fundamental Theorem of Calculus (FTC)

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Fundamental Theorem of Calculus 1

FTC Part 1
Let f(x)f\left(x\right) be a continuous function on[a, b]\left[a,\ b\right], then ddx∫axf(t)dt=f(x)\displaystyle \frac{d}{dx}\int_a^xf\left(t\right)dt=f\left(x\right)
  • Use this if the question involves both differentiation and integration side by side
  • It shows that differentiation and integration "cancel" each other
  • On the exam, use this short-cut formula for any FTC I questions: ddx∫g(x)h(x)f(t)dt=f(h)h′−f(g)g′\displaystyle \frac{d}{dx}\int_{g(x)}^{h(x)}f\left(t\right)dt=f\left(h\right)h'-f\left(g\right)g'

Example
Find ddx∫4xπ(t2−1)3dt\displaystyle \frac{d}{dx}\int_{4x}^{\pi}\left(t^2-1\right)^3dt.
Since we see differentiation and integration together, let's use the short-cut formula for FTC I:
=((π)2−1)3×0−((4x)2−1)3×4=\left(\left(\textcolor{orange}\pi\right)^2-1\right)^3\times\textcolor{orange}0-\left(\left(\textcolor{lime}{4x}\right)^2-1\right)^3\times\textcolor{lime}{4}
=−4(16x2−1)3=-4\left(16x^2-1\right)^3

Practice: FTC I

If g(x)=∫x2excos⁡t dt\displaystyle g\left(x\right)=\int_{x^2}^{e^x}\cos t\ dt, find g′(0)g'\left(0\right).

Practice: Application of FTC I

Given that lim⁡x→0∫0x(t2−1)dt=0, \displaystyle \lim_{x\rightarrow0}\int_0^x\left(t^2-1\right)dt=0,\ determine lim⁡x→0∫0x(t2−1)dtsin⁡x\displaystyle \lim_{x\rightarrow0}\frac{\displaystyle \int_0^x\left(t^2-1\right)dt}{\sin x}.

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Fundamental Theorem of Calculus 2

FTC Part 2
Suppose that F(x)F\left(x\right) is the antiderivative of f(x)f\left(x\right), then ∫abf(x)dx=F(b)−F(a)\displaystyle \int_a^bf\left(x\right)dx=F(b)-F(a)
  • Use this if the question asks you to evaluate a definite integral
  • It shows that the definite integral is a number
  • You need to find the antiderivative, then sub in the upper bound −- sub in the lower bound

Example
Evaluate ∫141−x3x2 dx\displaystyle \int_1^4\frac{1-x^3}{x^2}\ dx.
First simplify the expression:
=∫14 [1x2−x]dx=\displaystyle \int_{\color{orange}1}^{\color{lime}4}\ \left[\frac{1}{x^2}-x\right]dx
=∫14[x−2−x]dx\displaystyle =\int_{\color{orange}1}^{\color{lime}{4}}\left[x^{-2}-x\right]dx

Find the antiderivative:
=[x−1−1−x22]14\displaystyle =\left[\frac{x^{-1}}{-1}-\frac{x^2}{2}\right]_{\color{orange}1}^{\color{lime}4}
=[−1x−x22]14\displaystyle =\left[-\frac{1}{x}-\frac{x^2}{2}\right]_{\color{orange}1}^{\color{lime}4}
=[−14−422]−(−11−122)\displaystyle =\left[-\frac{1}{\color{lime}4}-\frac{\color{lime}{4}^2}{2}\right]-\left(-\frac{1}{\color{orange}1}-\frac{\color{orange}{1}^2}{2}\right)
=−274=-\frac{27}{4}

Practice: Definite Integral

Evaluate ∫0319+x2+2xln⁡2 dx\displaystyle \int_0^3\frac{1}{9+x^2}+2^x\ln2\ dx.

Practice: Definite Integral

Given that bothf(x)f\left(x\right) andf′(x)f'\left(x\right) are continuous everywhere, iff(1)=10f\left(1\right)=10and ∫−31f′(x)dx=12\displaystyle \int_{-3}^1f'\left(x\right)dx=12, find f(−3)f\left(-3\right).

Practice: Definite Integral

Evaluate ∫−20∣x2−1∣dx\displaystyle \int_{-2}^0\left|x^2-1\right|dx

Extra Practice
Fundamental theorem of calculus
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.
Fundamental theorem of calculus
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.
Fundamental theorem of calculus
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.
Application of Fundamental Theorem of Calculus
For what values of xx is the function f(x)=∫x22(t−1)dtf\left(x\right)=\int_{x^2}^2\left(t-1\right)dt decreasing?
Fundamental theorem of calculus
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 ?
Fundamental theorem of calculus
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.