Calculus 1: Definite Integrals and Fundamental Theorem
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All Calculus 1LimitsDefinition of the DerivativeProduct and Quotient RulePower Rule and Basic DerivativesDerivatives of Trig FunctionsExponential and Logarithmic FunctionsChain RuleInverse and Hyperbolic Trig DerivativesImplicit DifferentiationRelated Rates ProblemsLogarithmic DifferentiationGraphing and Critical PointsOptimization ProblemsIndeterminate Forms and l'Hospital's RuleLinear Approximation and DifferentialsNewton Raphson MethodIndefinite IntegralsU SubstitutionDefinite Integrals and Fundamental TheoremApplications of Integration
Evaluate the definite integral below
##\displaystyle\int_{-2}^2 \ {x^2 \cos{(\frac{x^3}{8})}} \ dx##
##\displaystyle\int_{-2}^2 \ {x^2 \cos{(\frac{x^3}{8})}} \ dx##
Evaluate the following definite integral
##\displaystyle\int_0^4 \ x \sqrt{x^2 + 9} \ dx##
##\displaystyle\int_0^4 \ x \sqrt{x^2 + 9} \ dx##
Find the area under the curve over the interval ##[0,4]##
##f(x) = x^2 + 1##
##f(x) = x^2 + 1##
Find the area under the curve over the interval ##[1,4]##
##f(x) = \frac{2}{x}##
##f(x) = \frac{2}{x}##
Evaluate the integral
##\displaystyle\int_0^2 (2x - 2x^2) \ dx##
##\displaystyle\int_0^2 (2x - 2x^2) \ dx##
Prove the fundamental theorem of calculus
Evaluate the integral ##\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (2 - \csc^2{x}) \ dx##
Find the area the region bounded by:
##y = 1 + \sqrt[3]{x}##
##x = 0##
##x = 8##
##y = 0##
##y = 1 + \sqrt[3]{x}##
##x = 0##
##x = 8##
##y = 0##
Compute the definite integrals
##\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \tan (x) \ dx## and ##\displaystyle\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} \tan (x) \ dx##
##\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \tan (x) \ dx## and ##\displaystyle\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} \tan (x) \ dx##